According to the Mean Value Theorem, there exists at least one value c in the interval (-5, 9) such that f'(c) = -23.71. The approximate value of c is -24.14.
To check if the Mean Value Theorem (MVT) can be applied to the function f(x) = 2x² - 3x² - 72x + 6 on the interval [-5, 9], we need to verify two conditions:
The function f(x) must be continuous on the closed interval [a, b].The function f(x) must be differentiable on the open interval (a, b).Let's check these conditions:
Continuity: The function f(x) is a polynomial, and polynomials are continuous for all values of x. Therefore, f(x) is continuous on the interval [-5, 9].Differentiability: The function f(x) is also a polynomial, and polynomials are differentiable for all values of x. Therefore, f(x) is differentiable on the interval (-5, 9).Since both conditions are satisfied, we can conclude that the Mean Value Theorem applies to f(x) on the interval [-5, 9].
According to the Mean Value Theorem, there exists at least one value c in the interval (-5, 9) such that the derivative of f evaluated at c is equal to the average rate of change of f over the interval [-5, 9].
To find the value(s) of c, we need to find the derivative of f(x) and set it equal to the average rate of change.
f(x) = 2x² - 3x² - 72x + 6
Taking the derivative:
f'(x) = 4x - 6x - 72
Simplifying:
f'(x) = -2x - 72
Now, we calculate the average rate of change of f over the interval [-5, 9]:
Average rate of change = (f(b) - f(a)) / (b - a)
= (f(9) - f(-5)) / (9 - (-5))
= (2(9)² - 3(9)² - 72(9) + 6 - [2(-5)² - 3(-5)² - 72(-5) + 6]) / (9 - (-5))
= (162 - 243 - 648 + 6 - 50 + 75 + 360 + 6) / 14
= -332 / 14
= -23.71
We need to find the value(s) of c such that f'(c) = -23.71.
Solving -2c - 72 = -23.71, we find:
-2c = -23.71 + 72
-2c = 48.29
c ≈ -24.14
Therefore, according to the Mean Value Theorem, there exists at least one value c in the interval (-5, 9) such that f'(c) = -23.71. The approximate value of c is -24.14.
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4500-p² 4 The demand equation for a product is found to be a = price of the product in dollars and q is the quantity. a. Find the price elasticity of demand when the price is $40. b. Is the demand el
The demand equation for a product is a function that represents the relationship between the price of a product and the quantity demanded by consumers. The price elasticity of demand when the price is $40 is E= (40/Q) (dQ/dP) = (40/Q) (q/40) = 1 Therefore, demand is unit elastic.
The price elasticity of demand measures the responsiveness of the quantity demanded of a product to a change in its price.
It is a crucial concept in economics, particularly in understanding how consumers react to changes in prices.
To answer this question, we use the formula for price elasticity of demand:
E= (P/Q) (dQ/dP) where E is the elasticity,
P is the price of the product, Q is the quantity demanded, and
dQ/dP is the derivative of the quantity demanded with respect to the price.
Given the demand equation,
a = price of the product in dollars and q is the quantity.
Therefore, we can rewrite the equation as follows:
a = Pq Taking the derivative of both sides, we get:
da/dP
= q + P (dq/dP) Solving for dq/dP,
we get: dq/dP
= (da/dP - q)/P
Plugging in the values, we get:
dq/dP
= (1q - 0)/40
= q/40
Hence, the price elasticity of demand when the price is $40 is
E= (40/Q) (dQ/dP)
= (40/Q) (q/40)
= 1
Therefore, demand is unit elastic.
The demand is unit elastic if the percentage change in quantity demanded is equal to the percentage change in price.
Therefore, a change in price will lead to an equal change in quantity demanded.
If the elasticity is greater than 1, the demand is elastic.
If the elasticity is less than 1, the demand is inelastic.
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Jacobi wants to install an underground sprinkler system in her backyard the backyard is rectangular with side length 17 m and 26 m .the water pipe will run diagonally across the yard about how many metres of water pipe does Jacobi need .
The length of the pipe required would be 31.06 meters
The length of the pipe is the hypotenus of the triangle formed :
hypotenus = √opposite² + adjacent²substituting the values into our equation:
length of pipe = √17² + 26²
length of pipe = √965 = 31.06
Therefore, the length of the pipe needed is 31.06 meters
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How does sample size affect determinations of statistical significance? The _________ the sample, the ________.
a. larger; greater probability that the variable has an effect
b. smaller; greater probability that the variable has an effect
c. larger; the more confident you can be in your decision to reject or retain the null hypothesis
d. smaller; the more confident you can be in your decision to reject or retain the null hypothesis
The larger; the sample, the greater probability that the variable has an effect. The correct option is (a).
When it comes to determining statistical significance, a larger sample size increases the statistical power of the analysis.
This means that with a larger sample size, there is a greater probability of detecting a true effect or relationship between variables.
This is because a larger sample size provides more information and reduces the impact of random variability.
Option (a) correctly identifies that a larger sample size leads to a greater probability that the variable has an effect. With a larger sample size, the analysis has more statistical power to detect and accurately estimate the effects or relationships being investigated.
A larger sample size also increases the precision of the estimates and reduces the sampling error, making the results more reliable and representative of the population. It allows for more accurate inference and increases the confidence in the findings.
Therefore, option (c) is also partially correct, as a larger sample size provides more confidence in the decision to reject or retain the null hypothesis.
In summary, a larger sample size improves the ability to detect effects and increases the confidence in the statistical analysis and decision-making process. The correct option is (a).
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Let Y₁,..., Yn N(μ,0²). State the sampling distribution of Y = n=¹_₁ Y₁. -1 i=1 n1, Σ (Υ; – Υ)2. State the sampling distribution of S² = State the mean and variance of Y and S².
1. The sampling distribution of Y is a normal distribution with mean nμ and variance nσ².
2. The mean of the sampling distribution of S² is σ², and the variance is 2σ⁴ / (n-1).
In the given notation, Y₁, Y₂, ..., Yₙ are independent and identically distributed (i.i.d.) random variables following a normal distribution with mean μ and variance σ².
1. Sampling Distribution of Y = ∑(i=1 to n) Yᵢ:
The random variable Y represents the sum of n independent normal random variables. The sampling distribution of Y is also a normal distribution. The mean of the sampling distribution of Y can be obtained by the linearity of expectation:
E(Y) = E(∑(i=1 to n) Yᵢ) = ∑(i=1 to n) E(Yᵢ) = ∑(i=1 to n) μ = nμ
The variance of the sampling distribution of Y can be obtained by the linearity of variance:
Var(Y) = Var(∑(i=1 to n) Yᵢ) = ∑(i=1 to n) Var(Yᵢ) = ∑(i=1 to n) σ² = nσ²
Therefore, the sampling distribution of Y is a normal distribution with mean nμ and variance nσ².
2. Sampling Distribution of S²:
The random variable S² represents the sample variance calculated from a sample of n observations. The sampling distribution of S² follows a chi-square distribution with (n-1) degrees of freedom.The mean of the sampling distribution of S² is given by:
E(S²) = σ²
The variance of the sampling distribution of S² is given by:
Var(S²) = 2σ⁴ / (n-1)
Therefore, the mean of the sampling distribution of S² is σ², and the variance is 2σ⁴ / (n-1).
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A contour map is shown for a function f(x,y) on the rectangle R=[−3,6]×[−1,4]. a. Use the midpoint rule with m=2 and n=3 to estimate the value of ∬Rf(x,y)dA. b. Estimate the average value of the function f(x,y). fave≈ Hint
a. The estimated value of ∬Rf(x,y)dA is 105
b. The estimated average value of the function f(x, y) is 7.
a. The rectangle R=[−3,6]×[−1,4] is divided into m = 2 subintervals along the x-axis and n = 3 subintervals along the y-axis. Therefore, each subinterval has a width of Δx = (6 - (-3))/2 = 9/2 and a height of Δy = (4 - (-1))/3 = 5/3.
We can calculate the midpoint of each subrectangle using the formula:
[tex]x_i = x_min + (i - 0.5) * \Delta x\\y_j = y_min + (j - 0.5) * \Delta y[/tex]
where i = 1, 2, ..., m and j = 1, 2, ..., n.
Using the midpoint rule, the estimate of the double integral is given by:
∬Rf(x,y)dA ≈ Δx * Δy * ∑∑[tex]f(x_i, y_j)[/tex]
where the double summation is taken over all the midpoints (x_i, y_j) of the subrectangles.
Calculate the midpoints of the subrectangles.
[tex]x_1 = -3 + (1 - 0.5) * (9/2) = -3 + 4.5 = 1.5\\x_2 = -3 + (2 - 0.5) * (9/2) = -3 + 9 = 6\\y_1 = -1 + (1 - 0.5) * (5/3) = -1 + (1/2) * (5/3) = -1 + 5/6 = -1/6\\y_2 = -1 + (2 - 0.5) * (5/3) = -1 + (3/2) * (5/3) = -1 + 5/2 = 9/2\\y_3 = -1 + (3 - 0.5) * (5/3) = -1 + (5/2) * (5/3) = -1 + 25/6 = 19/6[/tex]
Evaluate the function at each midpoint.
[tex]f(x_1, y_1) = 2\\f(x_1, y_2) = -1\\f(x_1, y_3) = 0\\f(x_2, y_1) = 1\\f(x_2, y_2) = 3\\f(x_2, y_3) = 2[/tex]
∬Rf(x,y)dA ≈ Δx * Δy * ∑∑[tex]f(x_i, y_j)[/tex]
= (9/2) * (5/3) * (2 + (-1) + 0 + 1 + 3 + 2)
= (9/2) * (5/3) * 7
= 15 * 7
= 105
b. To estimate the average value of the function f(x, y), we can divide the double integral by the area of the rectangle R, which is A = Δx * Δy * m * n.
The average value is then given by:
f_ave ≈ (∬Rf(x,y)dA) / A
Now let's perform the calculations:
Step 1: Calculate the area of the rectangle.
A = Δx * Δy * m * n
= (9/2) * (5/3) * 2 * 3
= 15
Step 2: Calculate the average value.
f_ave ≈ (∬Rf(x,y)dA) / A
= 105 / 15
= 7
Therefore, the estimated value of ∬Rf(x,y)dA is 105 and the estimated average value of the function f(x, y) is 7.
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The graph shows the function f(x) = |x – h| + k. What is the value of k?
The calculated value of k is -2.5
How to determine the value of k?From the question, we have the following parameters that can be used in our computation:
The graph
(see attachment)
Also, we have
f(x) = |x - h| + k
From the graph, we have the vertex to be
(h, k) = (1, -2.5)
By comparison, we have
k = -2.5
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