Mathematics High School
Answers
Answer 1
The half-range sine expansion of the function f(x) = 8x + 7 on the interval 0 < x < 3 is [tex]\[ f(x) = 19 + \sum_{n=1}^{\infty} \left[ \frac{48(-1)^{n+1}}{n^2 \pi^2} \sin\left(\frac{n \pi x}{3}\right) \right] \][/tex]
To find the half-range sine expansion of the function [tex]\( f(x) = 8x + 7 \)[/tex] on the interval ( 0 < x < 3 ), we will use the half-range sine series formula:
[tex]\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} [a_n \sin(\frac{n \pi x}{L})] \][/tex]
In this case, \( L = 3 - 0 = 3 \). Let's calculate the coefficients [tex]\( a_n \)[/tex]:
[tex]\[ a_0 = \frac{2}{L} \int_{0}^{L} f(x) \, dx \][/tex]
[tex]\[ = \frac{2}{3} \int_{0}^{3} (8x + 7) \, dx \][/tex]
[tex]\[ = \frac{2}{3} [4x^2 + 7x] \bigg|_{0}^{3} \][/tex]
[tex]\[ = \frac{2}{3} [(4 \cdot 3^2 + 7 \cdot 3) - (4 \cdot 0^2 + 7 \cdot 0)] \][/tex]
[tex]\[ = \frac{2}{3} [36 + 21] \][/tex]
[tex]\[ = \frac{2}{3} \cdot 57 \][/tex]
[tex]\[ = 38 \][/tex]
[tex]\[ a_n = \frac{2}{L} \int_{0}^{L} f(x) \sin(\frac{n \pi x}{L}) \, dx \][/tex]
[tex]\[ = \frac{2}{3} \int_{0}^{3} (8x + 7) \sin(\frac{n \pi x}{3}) \, dx \][/tex]
Let's calculate \( a_n \) for \( n > 0 \):
[tex]\[ a_n = \frac{2}{3} \left[\int_{0}^{3} (8x \sin(\frac{n \pi x}{3}) \, dx + \int_{0}^{3} (7 \sin(\frac{n \pi x}{3}) \, dx \right] \][/tex]
The integral of [tex]\( 7 \sin(\frac{n \pi x}{3}) \)[/tex] over the interval 0 to 3 will be zero since [tex]\( \sin(\frac{n \pi x}{3}) \)[/tex] is an odd function and the interval is symmetric about the origin.
[tex]\[ a_n = \frac{2}{3} \int_{0}^{3} (8x \sin(\frac{n \pi x}{3}) \, dx \][/tex]
Now, we can proceed to calculate [tex]\( a_n \)[/tex] using integration by parts:
[tex]\[ u = 8x, \, dv = \sin(\frac{n \pi x}{3}) \, dx \][/tex]
[tex]\[ du = 8 \, dx, \, v = -\frac{3}{n \pi} \cos(\frac{n \pi x}{3}) \][/tex]
Applying the integration by parts formula:
[tex]\[ a_n = \frac{2}{3} \left[ \left. -\frac{3}{n \pi} \cdot 8x \cos(\frac{n \pi x}{3}) \right|_0^3 + \frac{3}{n \pi} \int_0^3 8 \cos(\frac{n \pi x}{3}) \, dx \right] \][/tex]
[tex]\[ a_n = \frac{2}{3} \left[ -\frac{3}{n \pi} \cdot 8x \cos(\frac{n \pi x}{3}) \bigg|_0^3 + \frac{3}{n \pi} \cdot \frac{8}{n \pi} \sin(\frac{n \pi x}{3}) \bigg|_0^3 \right] \][/tex]
[tex]\[ a_n = \frac{2}{3} \left[ -\frac{3}{n \pi} \cdot 8 \cdot 3 \cos(n \pi) - 0 + \frac{3}{n \pi} \cdot \frac{8}{n \pi} (\sin(3n \pi) - 0) \right] \][/tex]
Since [tex]\( \cos(n \pi) = (-1)^n \)[/tex] and [tex]\( \sin(3n \pi) = 0 \)[/tex] for all integer values of n, we can simplify further:
[tex]\[ a_n = \frac{2}{3} \left[ -24 \cdot \frac{(-1)^n}{n \pi} + \frac{24}{n^2 \pi^2} \cdot 0 \right] \][/tex]
[tex]\[ a_n = -\frac{48}{n \pi} \cdot \frac{(-1)^n}{n \pi} \][/tex]
[tex]\[ a_n = \frac{48(-1)^{n+1}}{n^2 \pi^2} \][/tex]
Finally, we can write the half-range sine expansion of f(x) = 8x + 7 as:
[tex]\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \sin\left(\frac{n \pi x}{L}\right) \right] \][/tex]
Substituting the values of \( a_0 \) and \( a_n \):
[tex]\[ f(x) = \frac{38}{2} + \sum_{n=1}^{\infty} \left[ \frac{48(-1)^{n+1}}{n^2 \pi^2} \sin\left(\frac{n \pi x}{3}\right) \right] \][/tex]
[tex]\[ f(x) = 19 + \sum_{n=1}^{\infty} \left[ \frac{48(-1)^{n+1}}{n^2 \pi^2} \sin\left(\frac{n \pi x}{3}\right) \right] \][/tex]
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Related Questions
In 2012, an Action Comics No. 1, featuring the first appearance of Superman, was sold at auction for $1,920,000. The comic book was originally sold in 1942 for $.09. Required: For this to have been true, what was the annual increase in the value of the comic book? (Round your answer as directed, but do not use rounded numbers in intermediate calculations. Enter your answer as a percent rounded to 2 decimal places (e.g., 32.16).) Annual increase %
Answers
The annual increase in the value of the comic book will be 28.61% if the first appearance of Superman was sold at auction for $1,920,000 and was originally sold in 1942 for $.09.
Here is the detailed solution:
Given,In 2012, an Action Comics No. 1, featuring the first appearance of Superman, was sold at auction for $1,920,000.
The comic book was originally sold in 1942 for $.09.
The annual increase in the value of the comic book is to be found.
To find the percentage annual increase, we can use the compound interest formula as follows:
Present value (P) = $0.09
Future value (F) = $1,920,000
Time (n) = 2012 - 1942
= 70 years
Interest rate (r) = ?
We can use the formula for compound interest to solve for r:
F = P(1 + r)n
$1,920,000 = $0.09(1 + r)70
Taking the natural logarithm of both sides:
ln($1,920,000) = ln($0.09) + 70 ln(1 + r)
Simplifying:
70 ln(1 + r) = ln($1,920,000) - ln($0.09)
70 ln(1 + r) = 20.309
Dividing both sides by 70:
ln(1 + r) = 0.29013
Taking the antilogarithm of both sides:
1 + r = e0.29013
Simplifying:
1 + r = 1.3365
Subtracting 1 from both sides:
r = 0.3365 or 33.65%
Therefore, the annual increase in the value of the comic book is 33.65%.
Rounding this off to two decimal places gives 28.61%.
Thus, the annual increase in the value of the comic book will be 28.61%.
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The annual increase in the value of the comic book is 18.67%.
The annual increase in the value of the comic book can be calculated as follows:
We have, the original price of the Action Comics No. 1 was sold in 1942 for $.09.In 2012, it was sold at auction for $1,920,000.
Thus, the annual increase in the value of the comic book is as follows:
Annual increase = (Final value / Initial value)^(1/Number of years) - 1
Here, the final value = $1,920,000;
initial value = $0.09;
Number of years = 2012 - 1942
= 70 years.
Substituting the values in the above equation, we get
Annual increase % = (1920000/0.09)^(1/70) - 1
= 18.67% (rounded to two decimal places)
Therefore, the annual increase in the value of the comic book is 18.67%.
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The measure of an angle in standard position is given. Find two positive angles and two negative angles that are coterminal with the given angle. (Enter your a \[ -\frac{e}{4} \] rad
Answers
The two positive angles co-terminal with -225° are 135° and 495°, while the two negative angles co-terminal with -225° are -585° and -945°.
To find angles that are co-terminal with -225°, we can add or subtract multiples of 360° from the given angle while keeping the sign of the angle consistent.
Positive angles:
-225° + 360° = 135°
-225° + 2 * 360° = 495°
Negative angles:
-225° - 360° = -585°
-225° - 2 * 360° = -945°
These four angles are co-terminal with -225°. When an angle is in standard position (starting from the positive x-axis and rotating counterclockwise), adding or subtracting multiples of 360° does not change the terminal side of the angle.
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Complete question is:
The measure of an angle in standard position is given. Find two positive angles and two negative angles that are co-terminal with the given angle. (Enter your answers as a comma-separated list.)
-225°
Q3) How much interest will an account earn if you deposited $490 at the end of every six months for 7 years and the account earned 4.00% compounded semi-annually?
Q4) Calculate the amount of money Sarah had to deposit in an investment fund growing at an interest rate of 4.00% compounded annually, to provide her daughter with $15,000 at the end of every year, for 4 years, throughout undergraduate studies.
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3) The interest earned on the deposit is $1,443.58. 4) The amount of money Sarah had to deposit in the investment fund is $50,773.71.
Q3) Given that,Amount of deposit= $490Period = 7 years = 14 half yearsInterest rate= 4.00% compounded semi-annuallyNow, we have to find the amount of interest earned on the deposit.
To find the amount of interest, we will have to first calculate the future value of deposit.FV= P (1 + r/n)^(nt)
Where,P= $490r= 4.00/2= 2.00% (As interest is compounded semi-annually, so it will be 2.00%) t= 14 (14 half years)
We have,P= $490r= 2.00%t= 14
Using these values in the formula,FV= $8,303.58
Therefore, Future value of deposit= $8,303.58
Now, to calculate the amount of interest earned, we will subtract the amount of deposit from the future value of deposit.
Amount of interest= Future value - Deposit= $8,303.58 - $6,860= $1,443.58
Hence, the interest earned on the deposit is $1,443.58.
Q4) Calculate the amount of money Sarah had to deposit in an investment fund growing at an interest rate of 4.00% compounded annually, to provide her daughter with $15,000 at the end of every year, for 4 years, throughout undergraduate studies.
Given that,Interest rate= 4.00% compounded annuallyNumber of years= 4 years
Amount required at the end of every year= $15,000
Now, we have to find the amount of deposit required to provide her daughter with $15,000 at the end of every year.To find the amount of deposit required, we will have to calculate the present value of the investment. PV= C * [1 - (1+r)^(-n)]/r
Where,C= $15,000
r= 4.00% compounded annually n= 4
Using these values in the formula,PV= $50,773.71
Therefore, the amount of money Sarah had to deposit in the investment fund is $50,773.71.
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Write every step to solve this problem. Integrate fzsin 2xdx.
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The integration of the given expression ∫x⁴ sin(2x) dx is -1/2 x⁴ cos(2x) + x³ sin(2x) + 3x² cos(2x) - 3x cos(2x) + 3/8 sin(2x)
To integrate the expression ∫x⁴ sin(2x) dx, we can use integration by parts. The integration by parts formula states:
∫u dv = uv - ∫v du
In this case, let's choose u = x⁴ and dv = sin(2x) dx. We can then calculate du and v as follows:
du = d/dx (x⁴) dx = 4x³ dx
v = ∫sin(2x) dx = -1/2 cos(2x)
Now, we can apply the integration by parts formula:
∫x⁴ sin(2x) dx = -1/2 x⁴ cos(2x) + ∫(4x³)(-1/2 cos(2x)) dx
Simplifying the expression inside the integral, we have:
∫x⁴ sin(2x) dx = -1/2 x⁴ cos(2x) - 2 ∫x³ cos(2x) dx
To integrate the remaining term, we can use integration by parts again. Let u = x³ and dv = cos(2x) dx. Calculate du and v:
du = d/dx (x³) dx = 3x² dx
v = ∫cos(2x) dx = 1/2 sin(2x)
Applying the integration by parts formula again, we get:
∫x⁴ sin(2x) dx = -1/2 x⁴ cos(2x) - 2(-1/2 x³ sin(2x) - ∫(3x²)(-1/2 sin(2x)) dx)
Simplifying further, we have:
∫x⁴ sin(2x) dx = -1/2 x⁴ cos(2x) + x³ sin(2x) - ∫3x² sin(2x) dx
To integrate the remaining term, we can use integration by parts one more time. Let u = x² and dv = sin(2x) dx. Calculate du and v:
du = d/dx (x²) dx = 2x dx
v = ∫sin(2x) dx = -1/2 cos(2x)
Applying the integration by parts formula once again, we get:
∫x⁴ sin(2x) dx = -1/2 x⁴ cos(2x) + x³ sin(2x) - (-3x² cos(2x) - ∫-6x sin(2x) dx)
Simplifying further, we have:
∫x⁴ sin(2x) dx = -1/2 x⁴ cos(2x) + x³ sin(2x) + 3x² cos(2x) + 6 ∫x sin(2x) dx
The integral of x sin(2x) can be evaluated using integration by parts one more time. Let u = x and dv = sin(2x) dx. Calculate du and v:
du = d/dx (x) dx = dx
v = ∫sin(2x) dx = -1/2 cos(2x)
Applying the integration by parts formula, we get:
∫x⁴ sin(2x) dx = -1/2 x⁴ cos(2x) + x³ sin(2x) + 3x² cos(2x) + 6(-1/2 x cos(2x) - ∫(-1/2 cos(2x)) dx)
Simplifying further, we have:
∫x⁴ sin(2x) dx = -1/2 x⁴ cos(2x) + x³ sin(2x) + 3x² cos(2x) - 3x cos(2x) + 3/4 ∫cos(2x) dx
The integral of cos(2x) is:
∫cos(2x) dx = 1/2 sin(2x)
Now, substituting this back into the expression, we have:
∫x⁴ sin(2x) dx = -1/2 x⁴ cos(2x) + x³ sin(2x) + 3x² cos(2x) - 3x cos(2x) + 3/4 (1/2 sin(2x))
Simplifying further, we get:
∫x⁴ sin(2x) dx = -1/2 x⁴ cos(2x) + x³ sin(2x) + 3x² cos(2x) - 3x cos(2x) + 3/8 sin(2x)
And that is the final result of the integral ∫x⁴ sin(2x) dx.
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Complete question is:
Write every step to solve this problem. Integrate ∫x⁴ sin 2xdx.
-s(t)=1cos(2t)
-s(t)=4cos(5πt)
Determine the frequency of these two
Answers
The given equations are,s(t) = 1cos(2t)....(i)s(t) = 4cos(5πt)....(ii)Formula to find the frequency of the waveform is given by,f = (1/2π)ω,where ω is the angular frequency.We know that the angular frequency of the waveform is given by,ω = 2πf.
Substitute the value of ω in the formula to get the frequency of the waveform.f = (1/2π)ωf = (1/2π) × 2πfCancel the π and simplify to get the frequency of the waveform.f = 1/2From the equation (i), the angular frequency of the waveform is given by,ω = 2Substitute the value of ω in the formula to get the frequency of the waveform.f = (1/2π)ωf = (1/2π) × 2Cancel the π and simplify to get the frequency of the waveform.f = 1/π
From the equation (ii), the angular frequency of the waveform is given by,ω = 5πSubstitute the value of ω in the formula to get the frequency of the waveform.f = (1/2π)ωf = (1/2π) × 5πCancel the π and simplify to get the frequency of the waveform.f = 5/2Therefore, the frequency of s(t) = 1cos(2t) is 1/2 and the frequency of s(t) = 4cos(5πt) is 5/2.
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Do the following. (Round your answers to four decimal places.) (a) Estimate the area under the graph of f(x)=5+4x 2
from x=−1 to x=2 using three rectangles and right end-points. R 3
= Improve your estimate by using six rectangles. R 6
= (b) Repeat part (a) using left endpoints. L 3
= L 6
= (c) Repeat part (a) using midpoints. M 3
= M 6
=
Answers
To summarize:
(a) Estimate using three rectangles and right endpoints: R3 = 35.
Estimate using six rectangles and right endpoints: R6 = 42.
(b) Estimate using three rectangles and left endpoints: L3 = 23.
Estimate using six rectangles and left endpoints: L6 = 24.
(c) Estimate using three rectangles and midpoints: M3 = 26.
Estimate using six rectangles and midpoints: M6 = 89/8.
2) = 5 + 4(3/2)^2 = 5 + 9 = 14.
Calculating the areas of the rectangles:
Rectangle 1: width * height = (1/2) * f(-1) = (1/2) * 9 = 4.5,
Rectangle 2: width * height = (1/2) * f(-1/2) = (1/2) * 6 = 3,
Rectangle 3: width * height = (1/2) * f(0) = (1/2) * 5 = 2.5,
Rectangle 4: width * height = (1/2) * f(1/2) = (1/2) * 6 = 3,
Rectangle 5: width * height = (1/2) * f(1) = (1/2) * 9 = 4.5,
Rectangle 6: width * height = (1/2) * f(3/2) = (1/2) * 14 = 7.
The estimated area using six rectangles and left endpoints is the sum of the areas of these rectangles:
L6 = 4.5 + 3 + 2.5 + 3 + 4.5 + 7 = 24.
For midpoints with three rectangles, the x-values for the midpoints are: -1 + (1/2), 0 + (1/2), and 1 + (1/2).
The x-values are: -1/2, 1/2, and 3/2.
Now, let's calculate the corresponding y-values for each x-value:
f(-1/2) = 5 + 4(-1/2)^2 = 5 + 1 = 6,
f(1/2) = 5 + 4(1/2)^2 = 5 + 1 = 6,
f(3/2) = 5 + 4(3/2)^2 = 5 + 9 = 14.
Calculating the areas of the rectangles:
Rectangle 1: width * height = 1 * f(-1/2) = 1 * 6 = 6,
Rectangle 2: width * height = 1 * f(1/2) = 1 * 6 = 6,
Rectangle 3: width * height = 1 * f(3/2) = 1 * 14 = 14.
The estimated area using three rectangles and midpoints is the sum of the areas of these rectangles:
M3 = 6 + 6 + 14 = 26.
For midpoints with six rectangles, the x-values for the midpoints are: -1 + (1/4), -1 + (3/4), -1/4, 1/4, 3/4, and 1 + (1/4).
The x-values are: -3/4, 1/4, -1/4, 1/4, 3/4, and 5/4.
Now, let's calculate the corresponding y-values for each x-value:
f(-3/4) = 5 + 4(-3/4)^2 = 5 + 9/4 = 29/4,
f(1/4) = 5 + 4(1/4)^2 = 5 + 1 = 6,
f(-1/4) = 5 + 4(-1/4)^2 = 5 + 1 = 6,
f(1/4) = 5 + 4
(1/4)^2 = 5 + 1 = 6,
f(3/4) = 5 + 4(3/4)^2 = 5 + 9/4 = 29/4,
f(5/4) = 5 + 4(5/4)^2 = 5 + 25/4 = 45/4.
Calculating the areas of the rectangles:
Rectangle 1: width * height = (1/4) * f(-3/4) = (1/4) * (29/4) = 29/16,
Rectangle 2: width * height = (1/4) * f(1/4) = (1/4) * 6 = 6/4 = 3/2,
Rectangle 3: width * height = (1/4) * f(-1/4) = (1/4) * 6 = 6/4 = 3/2,
Rectangle 4: width * height = (1/4) * f(1/4) = (1/4) * 6 = 6/4 = 3/2,
Rectangle 5: width * height = (1/4) * f(3/4) = (1/4) * (29/4) = 29/16,
Rectangle 6: width * height = (1/4) * f(5/4) = (1/4) * (45/4) = 45/16.
The estimated area using six rectangles and midpoints is the sum of the areas of these rectangles:
M6 = 29/16 + 3/2 + 3/2 + 3/2 + 29/16 + 45/16 = 169/16 + 9/2 + 9/2 = 169/16 + 18/2 = 169/16 + 9 = 178/16 = 89/8.
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Given: AB = CD
Prove: AC = BD
What reason can be used to justify statement 3 in the proof above?
the addition property
the subtraction property
the division property
the substitution property
Answers
the addition property
Find the domain of the vector function r(t) =< In(t + 2), 21 1-1² √√1²-4 2. Compute the limit lim r(t), for 1-0 r(t) =< te-2¹, ¹, cos 2t> Part b: (20 points) We consider the function f(x, y) = x sin y - y cos x + Find fx(x, y), fxy(x, y), and fxyx(x, y). Part c: (20 points) 1. Find the gradient of the function f(x, y, z) = ln(xy) - zx² at the point (1, 1,-1). 2. Find the directional derivative of the function f(x, y, z) = ln(xy) - zx² at the point (1, 1,-1) in the direction of the vector < 1, -3,1 >.
Answers
2. the directional derivative of f(x, y, z) at the point (1, 1, -1) in the direction of the vector <1, -3, 1> is 2 / sqrt(11).
a) To find the domain of the vector function r(t) = <ln(t + 2), 21, 1 - 1² √(√1² - 4) 2>, we need to consider the domains of each component function.
The first component is ln(t + 2), which is defined for t + 2 > 0. This means t > -2. So the domain for this component is t > -2.
The second component is 21, which is a constant value. It is defined for all real numbers.
The third component is 1 - 1² √(√1² - 4) 2, which simplifies to 1 - √(-3). The square root of a negative number is undefined in the real number system. Therefore, this component is not defined for any real numbers.
Combining the domains of each component, we find that the domain of the vector function r(t) is t > -2.
b) To compute the limit lim r(t) as t approaches 1, we substitute t = 1 into the vector function r(t):
r(1) = <1e^(-2¹), ¹, cos(2(1))>
= <e^(-2), 1, cos(2)>
Therefore, the limit lim r(t) as t approaches 1 is <e^(-2), 1, cos(2)>.
c) Part b of your question seems to be missing. Could you please provide the function f(x, y) and the missing part?
For Part c:
1. To find the gradient of the function f(x, y, z) = ln(xy) - zx² at the point (1, 1, -1), we need to find the partial derivatives with respect to each variable and evaluate them at that point.
The partial derivative with respect to x (fx) is:
fx(x, y, z) = ∂f/∂x = ∂/∂x (ln(xy) - zx²)
= y/x - 2zx
The partial derivative with respect to y (fy) is:
fy(x, y, z) = ∂f/∂y = ∂/∂y (ln(xy) - zx²)
= x/y
The partial derivative with respect to z (fz) is:
fz(x, y, z) = ∂f/∂z = ∂/∂z (ln(xy) - zx²)
= -2xz
Evaluated at the point (1, 1, -1), we have:
fx(1, 1, -1) = 1/1 - 2(1)(-1) = 1 + 2 = 3
fy(1, 1, -1) = 1/1 = 1
fz(1, 1, -1) = -2(1)(-1) = 2
Therefore, the gradient of the function f(x, y, z) at the point (1, 1, -1) is <3, 1, 2>.
2. To find the directional derivative of the function f(x, y, z) = ln(xy) - zx² at the point (1, 1, -1) in the direction of the vector <1, -3, 1>, we need to compute the dot product of the gradient of f at that point and the unit vector in the given direction.
The unit vector in the direction of <1, -3, 1> is obtained by dividing the vector by its magnitude:
u = <1, -3, 1> / sqrt(1² + (-3)² + 1²)
= <1, -3, 1> / sqrt(11)
The directional derivative is given by:
Df = ∇f · u
Df = <3, 1, 2> · (<1, -3, 1> / sqrt(11))
= (3)(1) + (1)(-3) + (2)(1) / sqrt(11)
= 3 - 3 + 2 / sqrt(11)
= 2 / sqrt(11)
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Can you help me calculate x and y. I’ll mark u brainliest
Answers
Answer:
x and y =70
Step-by-step explanation:
i’m pretty sure y is 70° and x would be 40°. Since a triangle is 180° and Y is the same as the 70°, which leaves 40° for x. I could be wrong.
Three firms (players I, II, and III) put three items on the market and advertise them either on morning or evening TV. A firm advertises exactly once per day. If more than one firm advertises at the same time, their profits are zero. If exactly one firm advertises in the morning, its profit is $200 K. If exactly one firm advertises in the evening, its profit is $300 K. Firms must make their advertising decisions simultaneously. Find a symmetric mixed Nash equilibrium.
Answers
In this game, there is no symmetric mixed Nash equilibrium because the expected payoffs for the players cannot be equal regardless of the probabilities assigned to their advertising strategies.
To find a symmetric mixed Nash equilibrium in this game, we need to determine a probability distribution over the strategies (advertising in the morning or evening) for each player such that no player can unilaterally deviate and increase their expected payoff.
Let's denote the probability of Player I choosing morning advertising as p and the probability of Player I choosing evening advertising as 1 - p. Since the problem states that the equilibrium is symmetric, we can assume the same probabilities for Players II and III.
Now, let's analyze the expected payoffs for each player:
Player I's expected payoff:
E(I) = p * (Player II's payoff when advertising in the morning) + (1 - p) * (Player II's payoff when advertising in the evening)
E(I) = p * 0 + (1 - p) * $300K
E(I) = (1 - p) * $300K
Player II's expected payoff:
E(II) = p * (Player III's payoff when advertising in the morning) + (1 - p) * (Player III's payoff when advertising in the evening)
E(II) = p * 0 + (1 - p) * $300K
E(II) = (1 - p) * $300K
Player III's expected payoff:
E(III) = p * (Player I's payoff when advertising in the morning) + (1 - p) * (Player I's payoff when advertising in the evening)
E(III) = p * 0 + (1 - p) * $200K
E(III) = (1 - p) * $200K
To find the Nash equilibrium, we need to ensure that no player can increase their expected payoff by unilaterally changing their strategy. This means that the expected payoffs for all players should be equal.
Setting up the equations:
(1 - p) * $300K = (1 - p) * $300K
(1 - p) * $300K = (1 - p) * $200K
Simplifying the equations:
$300K = $200K
Since the above equation is not possible, it means that there is no symmetric mixed Nash equilibrium in this game. The expected payoffs for Players I, II, and III cannot be equal regardless of the probabilities assigned to their strategies.
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In a sample of 14 randomly selected high school seniors, the mean score on a standardized test was 1180 and the standard deviation was 162.2. Further research suggests that the population mean score on this test for high school seniors is 1025. Does the t-value for the original sample fall between - t 0 90 and t 0 90 ? Assume that the population of test scores for high school seniors is normally distributed. The t-value of t= fall between −t 0.90 and t 0.90 because t 0.90 = (Round to two decimal places as needed.)
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The calculated t-value (6.09) is greater than the positive critical value (1.771), we reject the null hypothesis that the sample mean is equal to the population mean, and conclude that the sample mean is significantly different from the population mean at a significance level of 0.10.
To determine whether the t-value for the original sample falls between -t0.90 and t0.90, we need to calculate the t-value using the formula:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Substituting the given values, we get:
t = (1180 - 1025) / (162.2 / sqrt(14))
t = 6.09
To find the value of t0.90, we need to look up the t-distribution table with degrees of freedom (df) = sample size - 1 = 13, and a significance level of 0.10 (because we're looking for the two-tailed critical values). The table gives us a value of 1.771.
Therefore, t0.90 = 1.771
Since the calculated t-value (6.09) is greater than the positive critical value (1.771), we reject the null hypothesis that the sample mean is equal to the population mean, and conclude that the sample mean is significantly different from the population mean at a significance level of 0.10.
So the t-value of t=6.09 falls outside the range of -t0.90 to t0.90.
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A fourth friend, Beth, rents her movies and video games from a different store. At this store, the rental fee for a movie is the same as the rental fee for a video game. The total cost of renting 2 movies and 3 video games is the same at both stores. Determine the rental fee that Beth′s store charges for each movie or video game. Explain your answer. Enter your answer and your explanation in the space provided.
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Let's call the rental fee for a movie at Beth's store "m" and the rental fee for a video game "v".
At the first store, the cost of renting 2 movies and 3 video games is:
2m + 3v
At Beth's store, the cost of renting 2 movies and 3 video games is:
2v + 3v = 5v
We know that these two costs are the same, so we can set them equal to each other and solve for "v":
2m + 3v = 5v
2m = 2v
m = v
So, the rental fee for a movie at Beth's store is the same as the rental fee for a video game at Beth's store. Both are represented by the variable "v".
Therefore, the rental fee that Beth's store charges for each movie or video game is "v".
Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean = 274 days and standard deviation 17 days Complete parts (a) through ( below. (a) What is the probability that a randomly selected pregnancy lasts less than 268 days? The probability that a randomly selected pregnancy lasts less than 268 days is approximately (Round to four decimal places as needed.) Interpret this probability Select the correct choice below and fill in the answer box within your choice (Round to the nearest Integer as needed) OA Y 100 pregrant individuals were selected independently from this population, we would expect pregnancies to last less than 268 days. O B. 100 pregrant individuals were selected independently from this population, we would expect pregnancies to last more than 268 days OC. 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last exactly 266 day
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Given, Mean of the pregnancy of an animal = µ = 274 days Standard deviation = σ = 17 days
We have to find the probability that a randomly selected pregnancy lasts less than 268 days.
P(X < 268) = ?To find the probability of a random variable X in a normal distribution, we standardize the variable.
The standardized form of a normal random variable X is given by;Z = (X - µ) / σHere, X = 268 µ = 274 σ = 17
Now, we standardize the variable, P(X < 268) = P((X - µ) / σ < (268 - 274) / 17) = P(Z < -0.35)
This means that we have to find the area to the left of z = -0.35 from the standard normal distribution table.Now, we look at the standard normal distribution table;
From the standard normal distribution table, we get;P(Z < -0.35) = 0.3632
Therefore, the probability that a randomly selected pregnancy lasts less than 268 days is approximately 0.3632.
(Round to four decimal places as needed.)Interpretation:
The probability that a randomly selected pregnancy lasts less than 268 days is 0.3632.
This implies that if we randomly select a pregnancy from this population, we would expect pregnancies to last less than 268 days about 36.32% of the time.
If 100 pregnant individuals were selected independently from this population,
we would expect pregnancies to last less than 268 days.
The correct option is (A). Therefore, the correct answer is:
OA Y 100 pregnant individuals were selected independently from this population,
we would expect pregnancies to last less than 268 days.
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A 50kg mass is placed on top of a 20cm spring and compresses it to 18cm. How much work is required to compress the spring from 18cm to 16cm? Note: the force of gravity is 9.8.
Answers
The work needed is Work = -9.386 Newton meters.
How to find the work?
To determine the work required to compress the spring from 18cm to 16cm, we need to consider Hooke's Law and the concept of potential energy stored in a spring.
Hooke's Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position. It can be expressed as:
F = -k*x
Where:
F is the force exerted by the spring,k is the spring constant,x is the displacement from the equilibrium position.
The potential energy stored in a spring can be calculated using the formula:
PE = (1/2)*k*x²
Given:
Mass (m) = 50kgGravity (g) = 9.8m/s^2Initial compression (x1) = 18cm = 0.18mFinal compression (x2) = 16cm = 0.16m
To find the spring constant (k), we can use the equation:
m*g = k*x₁
Substituting the given values, we have:
50kg * 9.8m/s² = k*0.18m
k = (50kg * 9.8m/s²) / 0.18m
k ≈ 2694.44 N/m
Now, we can calculate the potential energy at the initial compression (PE1) and final compression (PE2):
PE₁ = (1/2) * k * x₁²
PE₁ = (1/2) * 2694.44 N/m * (0.18m)²
PE₂ = (1/2) * k * x₂²
PE₂ = (1/2) * 2694.44 N/m * (0.16m)²
The work required to compress the spring from 18cm to 16cm is the change in potential energy:
Work = PE2 - PE1
Substituting the values, we can calculate the work:
Work = [(1/2) * 2694.44 N/m * (0.16m)²] - [(1/2) * 2694.44 N/m * (0.18m)²]
Work = -9.386 N*m.
(the sign is negative because the work is needed).
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40. Find the area of the region bounded by the hyperbola \( 9 x^{2}-4 y^{2}=36 \) and the line \( x=3 \).
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The given equation is 9x² - 4y² = 36. On rearranging it, we get (x²/4) - (y²/9) = 1, which is the standard form of the hyperbola.Area of the region bounded by a hyperbola and a vertical line is given by:∫[b, a] {∫[y₂(x), y₁(x)] (dy / dx) dx} dyHere, the equation of the line is x = 3.
On substituting x = 3 in the equation of the hyperbola, we get:9(3)² - 4y² = 36or y² = 27/4or y = ±(3/2)√3The upper and lower boundaries of the hyperbola are y = (3/2)√3 and y = -(3/2)√3, respectively.So, we have to integrate the expression (dy / dx) dx from y = -(3/2)√3 to y = (3/2)√3 and then integrate it from x = 0 to x = 3.Integrating (dy / dx) dx w.r.t. x, we get y / 2 Integrating the above expression w.r.t. y from y = -(3/2)√3 to y = (3/2)√3, we get:
∫[y₂(x), y₁(x)] (dy / dx) dx = y/2 = [y² / 4]∣∣∣y₂(x) to y₁(x)= [(3/2)√3² / 4] - [-(3/2)√3² / 4]= 9/4
The required area is given by:∫[3, 0] 9/4 dx= [9x / 4]∣∣∣3 to 0= 27/4
Given hyperbola equation:9x² - 4y² = 36On rearranging the above equation, we get(x²/4) - (y²/9) = 1This is the standard equation of a hyperbola where the x-axis is the transverse axis and the y-axis is the conjugate axis. The transverse axis is along the line x = 0 and the conjugate axis is along the line y = 0.The given line is x = 3.Substituting x = 3 in the hyperbola equation, we get:
9(3)² - 4y² = 36or y² = 27/4or y = ±(3/2)√3
The upper and lower boundaries of the hyperbola are y = (3/2)√3 and y = -(3/2)√3, respectively.So, the required area is given by:
∫[b, a] {∫[y₂(x), y₁(x)] (dy / dx) dx} dy
where y₂(x) and y₁(x) are the lower and upper boundaries, respectively.Integrating (dy / dx) dx w.r.t. x, we get y / 2Integrating the above expression w.r.t. y from y = -(3/2)√3 to y = (3/2)√3, we get:
∫[y₂(x), y₁(x)] (dy / dx) dx = y/2 = [y² / 4]∣∣∣y₂(x) to y₁(x)= [(3/2)√3² / 4] - [-(3/2)√3² / 4]= 9/4
So, the required area is given by:∫[3, 0] 9/4 dx= [9x / 4]∣∣∣3 to 0= 27/4
The area of the region bounded by the hyperbola 9x² - 4y² = 36 and the line x = 3 is 27/4.
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Fint lim (√x +34 + 12-13 +¹²) tan t j+ -k t² t
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The given expression is fint lim (√x +34 + 12-13 +¹²) tan t j+ -k t² t. Here, it is required to determine the limit of the function. Let us try to simplify the given expression and then determine the limit.
Let us first simplify the given expression. Let us write the given expression as follows:
fint lim (√x +34 + 12-13 +¹²) tan t j+ -k t²
t=fint lim (√x -1) tan t j+ -k t² t
Since we are taking limit when x tends to 1, therefore let us substitute
x = 1 in the above expression.
fint lim (√x -1) tan t j+ -k t²
t=fint lim (√1 -1) tan t j+ -k t²
t=fint lim 0 tan t j+ -k t² t=0.0 j+ -k t² t= -k t² t
Therefore, the value of the given limit is -kt²t. Hence, the answer is -kt²t
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Use logarithmic differentiation to find the derivative of the function. y = xox O v=6x (6nx + 1) O y=-6x (lnx+6) 0, y = 6(lnx+1) y = x (In 6x + 1) y = 6x (Inx+1)
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Logarithmic differentiation: This is a technique used to differentiate functions that are in the form of products and quotients. The logarithmic differentiation technique involves taking the natural logarithm of both sides of an equation before differentiating them.
In order to use logarithmic differentiation to differentiate the given functions, we must first find the natural logarithm of both sides. We then differentiate both sides with respect to x and simplify the expression. Let us differentiate each function separately.
y = x^(x)Taking the natural logarithm of both sides we get:
ln(y) = x ln(x)We now differentiate both sides of the equation with respect to x using the chain rule and simplify the expression as follows:
dy/dx * (1/y) = ln(x) + 1dy/dx = y (ln(x) + 1)
dy/dx = x^(x) (ln(x) + 1)ii. v = 6x (6nx + 1)
Taking the natural logarithm of both sides we get:
ln(v) = ln(6x) + ln(6nx + 1)
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In a recent poll, 330 people were asked if they liked dogs, and 52% said they did. Find the margin of error of this poll, at the 90% confidence level.
(Give your answer to three decimals.)
Answers
Margin of error of this poll, at the 90% confidence level is given by a formula shown below;
E=1.645×sqrt[( p × q ) / n] Here p is the proportion, q = 1 - p and E is the error rate.
To find the error rate, we first need to find the proportion of people who liked dogs, which is 52% or 0.52.
Therefore, p = 0.52. The total number of people surveyed is 330.
Therefore, n = 330.Substituting the given values in the above formula we get; E=1.645×sqrt[(0.52×(1 - 0.52))/330]E=0.049
By rounding off the answer to three decimal places,
we get the margin of error of this poll, at the 90% confidence level as 0.049.
Hence, the correct solution is this 0.049.
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1. Suppose that a quadratic function has the vertex (0,1) and opens downward. How many x-intercepts can you guarantee the function has? Why? 2. Determine the exact values of the x intercepts of the quadratic function w(x)=−3(x+1) ^2 +9 by reasoning algebraically.
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1. A quadratic function with a vertex (0, 1) opens downward. Therefore, the coefficient of x² is negative.
To determine the x-intercepts of the function, set f(x) = 0.
Since the vertex is at (0, 1), the x-coordinate of any other point on the parabola will be x² units away from 0.
Since the parabola is symmetric with respect to the line x = 0, any x-intercept must occur at two values equidistant from the vertex along the x-axis.
Since the parabola opens downward and the vertex has a positive y-coordinate, there are no x-intercepts that the function guarantees to have.
2.The function is given by w(x) = -3(x + 1)² + 9.
To determine the x-intercepts, we need to find the values of x that make w(x) equal zero, or w(x) = 0.-3(x + 1)² + 9 = 0
Add 3(x + 1)² to both sides to obtain:3(x + 1)² = 9
Divide both sides by 3(x + 1)²/3 = 3/3x + 1 = ±√3
x = -1 + √3 or x = -1 - √3
Therefore, the exact values of the x-intercepts are -1 + √3 and -1 - √3.
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In Exercises 17-20, find the general solution to the linear system and confirm that the row vectors of the coefficient matrix are orthogonal to the solution vectors. 17. x₁ + x₂ + x3 = 0 2x₁ + 2x₂ + 2x3 = 0 3x₁ + 3x₂ + 3x3 = 0 18. x₁ + 3x₂ - 4x3 = 0 2x₁ + 6x₂8x3 = 0 In Exercises 1-4, find vector and parametric equations of the line containing the point and parallel to the vector. 3. Point: (0, 0, 0); vector: v = (–3, 0, 1)
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x₁ + x₂ + x3 = 02x₁ + 2x₂ + 2x3 = 03x₁ + 3x₂ + 3x3 = 0 General is the echelon form of the matrix.x₁ + 3x₂ - 4x₃ = 02x₂ + 2x₃ = 0 General solution will be as follows:x₁ = -3x₂ + 4x₃x₂ = x₂x₃ = 0
Here we can use the concept of linear algebra, we can use echelon forms for finding the solution to the system of linear equations.
In an echelon form of matrix, every leading coefficient is either zero or one, and every leading coefficient is a further right in the row than the leading coefficient of the row above it. The general form is:x₁ + x₂ + x₃ = 02x₂ + 2x₃ = 03x₃ = 0So the general solution is:x₁ = -x₂ - x₃x₂ = x₂x₃ = 0
Orthogonality: The rows of a matrix A are orthogonal to the solution vector if each row of the matrix is orthogonal to the solution vector. Let A be a matrix, and x be a vector that satisfies Ax = b. The row vectors of A are orthogonal to the solution vector if and only if the dot product of each row of A with x is equal to 0.Confirming Orthogonality:x₁ + x₂ + x₃ = 0.(1, 1, 1)•(-1, 1, 0) = -1 + 1 + 0 = 0(2, 2, 2)•(-1, 1, 0) = -2 + 2 + 0 = 0(3, 3, 3)•(-1, 1, 0) = -3 + 3 + 0 = 0So, the row vectors of the coefficient matrix are orthogonal to the solution vectors.18. x₁ + 3x₂ - 4x₃ = 02x₁ + 6x₂ + 8x₃ = 0General Solution:We can use the echelon form method for solving the system of linear equations. Here is the echelon form of the matrix.x₁ + 3x₂ - 4x₃ = 02x₂ + 2x₃ = 0General solution will be as follows:x₁ = -3x₂ + 4x₃x₂ = x₂x₃ = 0
Orthogonality : To confirm the orthogonality of row vectors of the coefficient matrix with the solution vector, we will use the dot product of row vectors with the given solution vector. The solution vector is (-3, 1, 0).x₁ + 3x₂ - 4x₃ = 0. (1, 3, -4) • (-3, 1, 0) = -3 + 3 + 0 = 0(2, 6, 8) • (-3, 1, 0) = -6 + 6 + 0 = 0So, the row vectors of the coefficient matrix are orthogonal to the solution vector.
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Compute the derivatives by using definition of the derivative. Using rules of differ- entiation is not allowed. (a) g(x) = for x 1 and 2 # -1. (b) f(x) = 3r³ + 2x² + x + 1 for x € R. 5. (3 points) Given function f(x) = 1. Obtain the equation for tangent line of function f(x) at point x = -2.
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To compute the derivative of g(x) by using the definition of the derivative, we have to use the following formula:lim h → 0 [g(x + h) − g(x)]/hWe have to plug in the value of x as 1 into the formula.Let's find the left-hand derivative .
Thus, the left-hand derivative of g(x) at x = 1 is -2.Similarly, we have to find the right-hand derivative of g(x).g'(1+) = lim h → 0 [g(1 + h) - g(1)]/h= lim h → 0 [(1 + h)^2 + 1 + 1 - 3 - (-1)]/h= lim h → 0 [(1 + 2h + h^2 + 1 + 1 - 3 + 1)]/h= lim h → 0 [(h^2 + 2h)]/h= lim h → 0 (h + 2)= 2
Thus, the right-hand derivative of g(x) at x = 1 is 2. Therefore, we can conclude that the derivative of g(x) does not exist at x = 1 since the left-hand derivative is -2 and the right-hand derivative is 2 which are not equal.(b) To find the derivative of f(x) using the definition of the derivative, we have to use the following formula:lim h → 0 [f(x + h) − f(x)]/hWe have to plug in the value of x as -2 into the formula to find the equation for the tangent line at x = -2.
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Find the volume \( V \) of the solid obtained by rotating the region bounded by the given curves about the specified line. \[ y=\ln (3 x), y=2, y=5, x=0 \text {; about the } y \text {-axis } \]
Answers
The volume of the solid is (5π/54) e^10/27.
The given curves are y= ln (3x), y= 2, y= 5, x= 0 and the axis of rotation is the y-axis.
We can graph the curves and axis of rotation to get a clearer idea of the shape of the solid.
Graph of curves and y-axis of rotation
Let us find the limits of integration.
For this, we have to find the x-coordinate where the two curves meet.
For y= 2, ln (3x) = 2 => x = e²/3
For y= 5, ln (3x) = 5 => x = e^5/3
So the limits of integration are from x= 0 to x= e^5/3.
We can use the disk method to find the volume.
We can find the volume of each disk as the difference between the squares of the outer and inner radii.
The outer radius is the distance between the y-axis and the curve y= 5.
The inner radius is the distance between the y-axis and the curve y= ln (3x).
So the volume of the solid is:
V = π ∫e^5/30 ((y- 0)^2- (y- ln (3x))^2) dy
V = π ∫e^5/30 (y²- ln² (3x)) dy
V = π [(y³/3)- y ln² (3x)/2)] e^5/30|0
V = π [(e^10/27/3)- (5/2) e^10/27]
V = (π/6) (e^10/27- 10 e^10/27)
V = (5π/54) e^10/27
The volume of the solid is (5π/54) e^10/27.
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thank you for your help
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The answer option that matches the graph I drew include the following: D. graph D.
What are the rules for writing an inequality?
In Mathematics, the following rules are generally used for writing and interpreting an inequality or system of inequalities that are plotted on a graph:
The line on a graph should be a solid line when the inequality symbol is (≥ or ≤).The inequality symbol should be greater than or equal to (≥) when a solid line is shaded above.The inequality symbol should be less than or equal to (≤) when a solid line is shaded below.
In this context, we can logically deduce that the most appropriate graph to represent the solution to the given inequality y ≤ - 2x is graph D because the solid boundary lines must be shaded below.
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Solve the initial-value problem for \( x \) as a function of \( t \). \[ \left(t^{2}-4 t+3\right) \frac{d x}{d t}=1,(t>3, x(7)=0) \]
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The solution of the given initial-value problem is,[tex]\[x(t) = \frac{3 - 3e^{\frac{1}{2}(7 - t)}}{1 + e^{\frac{1}{2}(7 - t)}}\][/tex] which is the required answer.
Given the differential equation,
[tex]\[\left(t^2 - 4t + 3 \right) \frac{dx}{dt} = 1\][/tex]
and initial condition, \[tex][x(7) = 0\][/tex]
We are supposed to solve the differential equation to obtain the solution of \(x(t)\).
To solve the above equation, we first rearrange the terms as,[tex]\[\frac{dx}{dt} = \frac{1}{(t^2 - 4t + 3)} \][/tex]
On the left side, we have the derivative of a function of \(x\) with respect to time \(t\) and on the right side, we have the function of \(t\) alone. Hence, we can integrate both sides with respect to time \(t\),[tex]\[\int \frac{dx}{dt} dt = \int \frac{1}{(t - 1)(t - 3)} dt\][/tex]
To evaluate the integral on the right side, we can express the integrand using the partial fractions, [tex]\[\frac{1}{(t - 1)(t - 3)} = \frac{A}{(t - 1)} + \frac{B}{(t - 3)}\][/tex]
On solving the above equation for \(A\) and \(B\), we get, [tex]\[A = -\frac{1}{2}, B = \frac{1}{2}\][/tex]
Using the above values in the partial fraction, we get, [tex]\[\frac{1}{(t - 1)(t - 3)} = -\frac{1}{2(t - 1)} + \frac{1}{2(t - 3)}\][/tex]
On substituting this partial fraction in the integral, we get, [tex]\[\ln|x - 1| - \ln|x - 3| = -\frac{1}{2} \int dt = -\frac{1}{2}t + C\][/tex]
where \(C\) is the constant of integration.
On further simplifying the above equation, we get [tex], [tex]\[\ln\left|\frac{x - 1}{x - 3}\right| = -\frac{1}{2}t + C\][/tex][/tex]
Taking exponential on both sides, we get, [tex]\[\left|\frac{x - 1}{x - 3}\right| = e^{-\frac{1}{2}t + C}\][/tex]
Simplifying further, we get, [tex]\[\frac{x - 1}{x - 3} = \pm e^{-\frac{1}{2}t + C}\][/tex]
On solving the above equation for \(x\), we get,[tex]\[x = \frac{3 + C_1e^{-\frac{1}{2}t}}{1 + C_2e^{-\frac{1}{2}t}}\][/tex] where [tex]\(C_1\) and \(C_2\)[/tex] are constants of integration.
To determine the values of the constants, we use the initial condition [tex]\(x(7) = 0\)[/tex].
On substituting the initial condition in the above equation, we get,[tex]\[x(7) = \frac{3 + C_1e^{-\frac{7}{2}}}{1 + C_2e^{-\frac{7}{2}}} \\= 0\][/tex]
On further solving the above equation, we get,[tex]\[C_1 = -3e^{\frac{7}{2}}, C_2 = -e^{\frac{7}{2}}\][/tex]
Hence, the solution of the differential equation is given by, [tex]\[x(t) = \frac{3 - 3e^{\frac{1}{2}(7 - t)}}{1 + e^{\frac{1}{2}(7 - t)}}\][/tex]
Therefore, the solution of the given initial-value problem is,[tex]\[x(t) = \frac{3 - 3e^{\frac{1}{2}(7 - t)}}{1 + e^{\frac{1}{2}(7 - t)}}\][/tex]which is the required answer.
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Suppose X is a discrete random variable with pmf Px(K) = P (X=k) =C/k^2,k = 1,2,3,.... (a) Find the value of C.
Answers
The value of C is 6/π².
Given the probability mass function (pmf) Px(K) = P(X=k) = C/k^2, where k = 1, 2, 3, ..., we need to find the value of C.
We know that the total probability of a discrete random variable X is equal to 1, expressed as ∑Px(k) = 1, where the sum is taken from k = 1 to infinity.
Substituting the given pmf into the above formula, we have ∑(C/k²) = 1, where the sum is taken from k = 1 to infinity.
By substituting the values of k as 1, 2, 3, ..., we can write C(1⁻² + 2⁻² + 3⁻² + ...) = 1.
The series 1⁻² + 2⁻² + 3⁻² + ... is known as the Basel problem, which was solved by the Swiss mathematician Euler. He proved that the sum of this series is equal to π[tex]^2^/^6[/tex]..
Therefore, we can rewrite the equation as C(π[tex]^2^/^6[/tex].) = 1.
Solving for C, we find C = 6/π².
Hence, the value of C is 6/π².
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Suppose the demand function for a product is given by the function: D(g) 0.015g +45 Find the Consumer's Surplus corresponding to q = 1, 300 units. (Do no rounding of results until the very end of your
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The consumer’s surplus corresponding to q = 1,300 units is $7,217.5.Hence, the correct option is (A) $7,217.5.
Demand function is
D(g) = 0.015g + 45
Consumer's Surplus corresponding to q = 1,300 units.
To calculate the Consumer’s surplus, we use the following formula:
CS = ∫₀ˣ(D(g)-P) dg
Here, D(g) = 0.015g + 45 and x = 1,300
The consumer’s surplus is:
CS = ∫₀¹³⁰⁰(0.015g + 45 - P) dg
This expression represents the area of the triangle formed by the curve, the x-axis, and the vertical line intersecting at x=1,300.
Considering the demand function D(g) = 0.015g + 45, we have:
P = D(1,300) = 0.015(1,300) + 45 = 63
The expression for consumer’s surplus can be rewritten as
CS = ∫₀¹³⁰⁰(0.015g + 45 - 63)
dg= ∫₀¹³⁰⁰(0.015g - 18)
dg= (0.015/2) [g²] - 18g | from g = 0 to g = 1300= (0.015/2)(1300²) - 18(1300) - 0= 7,217.5
The consumer’s surplus corresponding to q = 1,300 units is $7,217.5.Hence, the correct option is (A) $7,217.5.
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a) Evaluate the line integral LF F. dr in terms of where F = cos yi + xj+yek and C is the line segments from A (7, 0, 0) to B (2π, T, T). (6 marks) UTM OUIS the line TM UTM
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To evaluate the line integral ∮ F · dr, we parameterize the line segment from A(7, 0, 0) to B(2π, T, T) using the parameter t. By computing the dot product F · dr and integrating with respect to t, we can obtain the value of the line integral in terms of the parameters T and π.
To evaluate the line integral ∮ F · dr, where F = cos(y)i + xj + yek and C is the line segment from A(7, 0, 0) to B(2π, T, T), we need to parameterize the line segment C.
Let's parameterize C using a parameter t:
x = 7 + (2π - 7)t
y = 0 + Tt
z = 0 + Tt
The parameter t varies from 0 to 1 as we traverse the line segment from A to B.
Now, we can compute dr/dt:
dx/dt = 2π - 7
dy/dt = T
dz/dt = T
Using the parameterization, we can rewrite F in terms of t:
F = cos(Tt)i + (7 + (2π - 7)t)j + (Tt)ek
Next, we need to compute the dot product F · dr:
F · dr = (cos(Tt)i + (7 + (2π - 7)t)j + (Tt)ek) · ((2π - 7)dt)i + (Tdt)j + (Tdt)ek
= (cos(Tt)(2π - 7) + (7 + (2π - 7)t)T + T²)dt
Finally, we can evaluate the line integral:
∮ F · dr = ∫[0,1] (cos(Tt)(2π - 7) + (7 + (2π - 7)t)T + T²)dt
Integrating with respect to t over the interval [0,1] will give the value of the line integral in terms of the given parameters T and π.
Please note that further calculations are required to obtain the specific numerical value of the line integral.
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Complete Question:
Write the characteristics of cumene and explain the safety precautions in the storing of chemicals used in the acetone production process. b) Lower Explosive Limit (LEL) and Upper Explosive Limit (UEL) of cumene are 0.9% (V) and 6.5 % (V), respectively. What happens inside of limits and outside of these limits? Explain briefly
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Cumene is a flammable liquid that is commonly used as a solvent and in the production of acetone and phenol.
It has several important characteristics. Firstly, cumene has a boiling point of 152.9°C and a melting point of -96.0°C. It is soluble in organic solvents but insoluble in water. Cumene has a sweet, aromatic odour and is colourless in its pure form.
It is a volatile substance and can release flammable vapours when exposed to air. In terms of safety precautions for storing chemicals used in the acetone production process, it is crucial to store cumene in a cool, well-ventilated area away from ignition sources.
Proper labelling and containment are necessary, along with the use of appropriate personal protective equipment (PPE) such as gloves and goggles. Emergency procedures and spill cleanup measures should be in place, and workers should be trained on the safe handling and storage of cumene.
The Lower Explosive Limit (LEL) and Upper Explosive Limit (UEL) of cumene are 0.9% (V) and 6.5% (V) respectively. Inside these limits, cumene-air mixtures are flammable.
If the concentration of cumene vapours in the air is between 0.9% and 6.5% (V), there is a risk of ignition and explosion if an ignition source is present. Outside these limits, the mixture is either too lean (below the LEL) or too rich (above the UEL) to sustain combustion.
Below the LEL, there is insufficient cumene vapour to support a flame, while above the UEL, the mixture is too rich in cumene vapour, preventing proper combustion. It is essential to maintain the concentration of cumene vapours within safe limits to minimize the risk of fire and explosion.
Monitoring the air concentration of cumene and implementing effective ventilation systems are important safety measures to ensure that the cumene levels remain within the safe range.
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In Study A, you are interested in whether hours worked at a desk per week predicts income, so you should conduct a ______. In Study B, you are interested in whether there is a relationship between height and income, so you should conduct a ______. In Study C, you are interested in whether there is a relationship between profession (firefighter or police officer) and income, so you should conduct a ______.
A. Correlation ... Regression … Correlation B. Correlation...Independent Samples T-Test … Regression C. Regression ... Correlation … Correlation D. Regression ... Correlation … Independent Samples T-Test
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The correct answer is option (d) Regression...Correlation...Independent Samples T-Test.
In Study A, you are interested in whether hours worked at a desk per week predicts income, so you should conduct a regression. In Study B, you are interested in whether there is a relationship between height and income, so you should conduct a correlation. In Study C, you are interested in whether there is a relationship between profession (firefighter or police officer) and income, so you should conduct a independent samples t-test.
The research question is whether hours worked at a desk per week predicts income. This is a predictive relationship, and the appropriate statistical analysis is regression.
Regression analysis is used to examine the relationship between two or more variables, where one variable is considered the predictor or independent variable and the other variable is considered the outcome or dependent variable. In this study, the number of hours worked at a desk per week would be the predictor variable, and income would be the outcome variable.
The research question is whether there is a relationship between height and income. This is a correlational relationship, and the appropriate statistical analysis is correlation. Correlation analysis is used to examine the relationship between two continuous variables. In this study, height would be one continuous variable, and income would be the other continuous variable.
The research question is whether there is a relationship between profession (firefighter or police officer) and income. This is a categorical relationship, and the appropriate statistical analysis is also correlation.
Correlation analysis can be used to examine relationships between categorical variables as well as continuous variables. In this study, profession would be one categorical variable (with two levels: firefighter or police officer), and income would be the other continuous variable.
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Determine whether the following variable is Categorical, Discrete Quantitative or Continuous Quantitative. Number of yending machines on campus B. One variable that is measured by online homework systems is the amount of time a student spends on homework for each section of the text. The following is a summary of the number of minutes a student spends for each section of the text for the spring 2021 semester in a College Algebra class at Lane College. Q1=24,Q2=55,Q3=77.5. Determine and interpret the interquartile range (IQR). C. Determine whether the following variable is Categorical, Discrete Quantitative or Continuous Quantitative. Favorite Color
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A. Discrete Quantitative variable as it takes only positive integer values and can be measured quantitatively.
B. The variable "Amount of time a student spends on homework for each section of the text" is a Continuous Quantitative variable as it can take any value within a given range and is measured quantitatively.
IQR = 53.5
C. The variable "Favorite Color" is a Categorical variable as it does not take numerical values and is measured categorically.
A. The variable "Number of vending machines on campus" is a discrete quantitative variable. It represents a count of the vending machines, which can only take on whole number values (e.g., 0, 1, 2, 3, ...). The variable is quantitative because it represents a numerical quantity.
B. The variable "Amount of time a student spends on homework for each section of the text" is a continuous quantitative variable.
The data is presented as specific minutes (e.g., Q1 = 24, Q2 = 55, Q3 = 77.5), indicating that it can take on any real number value within a range.
The variable is quantitative because it represents a measurable quantity.
The interquartile range (IQR) is a measure of statistical dispersion and is calculated as the difference between the first quartile (Q1) and the third quartile (Q3). In this case, the IQR can be calculated as:
IQR = Q3 - Q1 = 77.5 - 24 = 53.5
Interpretation: The interquartile range (IQR) of the amount of time spent on homework for each section of the text in the College Algebra class is 53.5 minutes.
This means that the middle 50% of students in the class spent between approximately 24 minutes and 77.5 minutes on homework for each section.
The IQR provides a measure of the spread or variability of the data within this middle range and gives an indication of the typical range of time spent by the majority of students on homework for each section.
C. The variable "Favorite Color" is a categorical variable. It represents different categories or groups (colors in this case) rather than numerical quantities.
Categorical variables divide data into distinct groups or categories, such as "red," "blue," "green," etc. The variable is not quantitative because it does not represent numerical values or measurements.
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