To answer these questions, we can use the properties of the normal distribution.
a) To find the proportion of houseflies with lengths between 6.3 and 6.5 millimeters, we need to calculate the area under the normal curve between these two values. We can use a standard normal distribution with mean 0 and standard deviation 1, and then convert back to the original distribution.
b) To find the proportion of houseflies with lengths greater than 6.5 millimeters, we need to calculate the area under the normal curve to the right of 6.5.
c) To estimate the number of houseflies in the sample with lengths greater than 6.5 millimeters, we can multiply the proportion found in part b) by the sample size (64).
d) To estimate the number of houseflies in the sample with lengths between 6.3 and 6.5 millimeters, we can subtract the estimate from part c) from the sample size (64).
e) The standard deviation of the distribution of X (sample mean) can be calculated by dividing the standard deviation of the original distribution (0.12 mm) by the square root of the sample size (√64).
f) The standard deviation of the distribution of Xtot (sample sum) can be calculated by multiplying the standard deviation of the original distribution (0.12 mm) by the square root of the sample size (√64).
g) To find the probability that 6.38 < X < 6.42 mm, we can calculate the area under the normal curve between these two values.
h) To find the probability that X tot > 415 mm, we need to calculate the area under the normal curve to the right of 415.
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Fill in the blanks to complete the following multiplication (enter only numbers): -2y (1-y+3y²) = − y³ + y²- y
The completed multiplication is -y³ + y² - y.
To complete the multiplication -2y(1-y+3y²), we need to distribute the -2y to each term inside the parentheses:
-2y x 1 = -2y
-2y x (-y) = 2y²
-2y x 3y² = -6y³
Adding up these terms, we get:
-2y + 2y² - 6y³
This demonstrates the concept of distributing or applying the distributive property in algebra. When we have a term multiplied by a polynomial, we need to multiply the term by each term in the polynomial and then combine the like terms, if any.
In this case, the term "-2y" is multiplied by each term in "(1-y+3y²)" to obtain the resulting expression.
Therefore, the completed multiplication is -y³ + y² - y.
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Consider the vector field F(x, y) = (-2xy, x² ) and the region R bounded by y = 0 and y = x(2-x)
(a) Compute the two-dimensional divergence of the field.
(b) Sketch the region
(c) Evaluate BOTH integrals in Green's Theorem (Flux Form) and verify that both computations match.
The given vector field F(x, y) = (-2xy, x²) is considered along with the region R bounded by y = 0 and y = x(2-x). The two-dimensional divergence of the field is computed.
(a) The two-dimensional divergence of the field F(x, y) = (-2xy, x²) is computed by taking the partial derivative of the first component with respect to x and the partial derivative of the second component with respect to y. The divergence is obtained as -2x.
(b) The region R bounded by y = 0 and y = x(2-x) is sketched. This region is the area between the x-axis and the curve y = x(2-x). It is a triangular region in the coordinate plane.
(c) Green's Theorem (Flux Form) is applied to evaluate two integrals. The first integral involves the line integral of the vector field F(x, y) = (-2xy, x²) over the boundary curve of the region R. The second integral involves the double integral of the divergence of F over the region R. Both integrals are computed, and it is verified that the values obtained from both computations match. This verifies the accuracy of Green's Theorem in this context.
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13. (5 points) Imagine that I asked you to construct a regular 24-gon inscribed in a circle and a regular 24-gon circumscribing a circle. I then asked you to find the areas of these two shapes. You worked very hard, and you found that the area of the smaller 24-gon was about 3.105, while the area of the larger 24-gon was about 3.160. Why might we be interested in this procedure and calculation? What is the historical significance? And why is a 24-gon a convenient shape?
In geometry, polygons are used as a building block for many geometric shapes. A regular polygon is a two-dimensional figure that has congruent sides and angles.
Regular polygons have a unique property that makes them special, they have sides that are all equal in length and angles that are all equal in measure.
Therefore, a regular polygon can be inscribed in a circle (all of its vertices lie on the circumference of the circle) and can be circumscribed around a circle (the circle passes through all of its vertices).
Inscribed polygonCircumscribed polygon 24-gon is a convenient shape since it is divisible by 2, 3, 4, 6, 8, and 12.
This property is because the number 24 has many factors, and it makes it easier to calculate the area of a regular 24-gon inscribed in a circle and a regular 24-gon circumscribing a circle.
Historical SignificanceThe ancient Greeks were interested in finding the exact areas of different shapes.
Archimedes was one of the ancient Greek mathematicians who developed an approach for finding the area of a circle.
In his work, he used a method called the "Method of Exhaustion," which involves approximating the area of a shape using inscribed and circumscribed polygons of a shape.
By using this method, Archimedes found an approximation for the value of pi.
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Research was conducted on the weight at birth of children from urban and rural women. The researcher suspects that there is a significant difference in the mean weight at birth of children between urban and rural women. The researcher selects independent random samples of mothers who gave birth from each group and calculates the mean weight at birth of children and standard deviations. The statistics are summarized in the table below. (a) Test whether there is a difference in the mean weight at birth of children between urban and rural women (use 5% significant level). (30 marks) (b) Assume that medical experts commonly believe that on average a new-born baby in urban areas weighs 3.5000 kg. Is it true that the observed mean weight at birth of children from sample urban mothers is greater than the predicted weight? (use 5% significant level). (20 marks)
(a) To test the difference in mean weight at birth between urban and rural women, a two-sample t-test can be used. The significance level of 5% implies that we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.
The t-test compares the means of the two samples, considering their respective sample sizes and standard deviations. By calculating the test statistic and comparing it to the critical value from the t-distribution with appropriate degrees of freedom, we can determine whether the observed difference is statistically significant.
(b) To test whether the observed mean weight at birth of children from sample urban mothers is greater than the predicted weight of 3.5000 kg, a one-sample t-test can be conducted. The null hypothesis (H₀) assumes that the mean weight is equal to or less than 3.5000 kg, while the alternative hypothesis (H₁) suggests that the mean weight is greater.
Similar to the previous test, the t-test calculates the test statistic using the sample mean, standard deviation, and sample size. By comparing the test statistic to the critical value from the t-distribution with appropriate degrees of freedom, we can determine whether the observed mean weight is significantly greater than the predicted weight.
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1) Differentiate. a) f(x)= 1 (cos(x5-5x)*
b) f(x) = sin-1(x3 - 3x)
The differentiation of the given functions are as follows; a) [tex]f(x) = 1cos(x5 - 5x) :[/tex]
[tex]df/dx = sin(x^5 - 5x)(5x^4 - 5)b) f(x)[/tex]
[tex]= sin-1(x3 - 3x) :[/tex]
[tex]f′(x) = (3x^2 - 3) / √(1 - (x^3 - 3x)^2).[/tex]
Differentiation of trigonometric functions The process of finding the derivative of a function is called differentiation. In mathematics, differentiation is a primary mathematical concept that has a variety of applications in various fields. It is applied to trigonometric functions as well. The trigonometric functions that are primarily differentiated include sine, cosine, tangent, cotangent, secant, and cosecant. Therefore, the differentiation of the given functions is as follows; a) [tex]f(x) = 1cos(x5 - 5x)[/tex] The given function is
[tex]f(x) = 1cos(x5 - 5x).[/tex] To find its derivative, we use the formula of the chain rule of differentiation:
[tex]`(f(g(x)))′ = f′(g(x))g′(x)`[/tex] Given that,
[tex]`f(x) = 1cos(x5 - 5x)`[/tex] Let
[tex]`u = (x^5 - 5x)`[/tex] So,
[tex]`f(x) = 1cosu`[/tex] Now differentiate `u` with respect to `x` and get `du/dx
[tex]= 5x^4 - 5`[/tex] Then
[tex]`df/dx = -sinu (du/dx)` But `cosu[/tex]
[tex]= cos(x^5 - 5x)`[/tex] Therefore, the differentiation of
[tex]f(x) = 1cos(x5 - 5x)[/tex] is given by
[tex]`df/dx = sin(x^5 - 5x)(5x^4 - 5)`b)[/tex]
[tex]f(x) = sin-1(x3 - 3x).[/tex]
The given function is [tex]f(x) = sin-1(x3 - 3x)[/tex] To find its derivative, we apply the formula of the chain rule of differentiation: [tex]`(f(g(x)))′ = f′(g(x))g′(x)`[/tex] Let
[tex]`u = x^3 - 3x`[/tex] and
[tex]`y = sin-1u`[/tex] Hence,
[tex]`y′ = dy/du * du/dx`[/tex] Differentiate `y` with respect to `u` and get
[tex]`dy/du = 1/√(1 - u^2)`[/tex] Differentiate `u` with respect to `x` and get
[tex]`du/dx = 3x^2 - 3`[/tex] Therefore,
[tex]`y′ = (1/√(1 - u^2)) * (3x^2 - 3) `[/tex] Hence, the differentiation of
[tex]f(x) = sin-1(x3 - 3x)[/tex] is given by
[tex]`f′(x) = (3x^2 - 3) / √(1 - (x^3 - 3x)^2)`[/tex] In conclusion, the differentiation of the given functions are as follows; a)
[tex]f(x) = 1cos(x5 - 5x)[/tex] :
[tex]df/dx = sin(x^5 - 5x)(5x^4 - 5)b) f(x)[/tex]
[tex]= sin-1(x3 - 3x)[/tex] :
[tex]f′(x) = (3x^2 - 3) / √(1 - (x^3 - 3x)^2).[/tex]
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Diversifying an investment portfolio increases the return to
risk ratio. Diversifying internationally heightens the benefits of
diversification. Explain why this is. Diversifying into frontier
and eme
A diversified-portfolio is important because of risk-reduction, smoother-returns, exploiting different opportunities, and risk-allocation.
A "Diversified-Portfolio" refers to an investment portfolio that contains a mix of different asset classes, industries, regions, and securities.
A diversified portfolio is important for several reasons, which are :
(i) Risk-reduction: Diversification helps to reduce the overall risk of investment portfolio. By spreading the investments across different asset classes, industries, regions, and securities, we can mitigate the impact of any individual investment performing poorly.
(ii) Smoother-returns: Diversification can lead to more stable and smoother investment returns over time. Different asset classes or investments tend to perform differently under various market conditions.
(iii) Exploiting different opportunities: By diversifying your portfolio, you can participate in various growth areas and potentially benefit from different economic cycles.
(iv) Risk-allocation: Diversification allows us to allocate the investment capital across different risk profiles based on your investment objectives and risk tolerance.
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The given question is incomplete, the complete question is
Why is it important to have a diversified portfolio?
Set up the definite integral required to find the area of the
region between the graph of y = 11 − x 2 and y = − 25 x + 165 over
the interval − 1 ≤ x ≤ 1
The integral we need to solve is:
[tex]\int\limits^1_{-1} {(- x^2 + 25 x - 154)} \, dx[/tex]
How to find the area between the curves?
Here we just need to integrate the difference between the two curves in the given region, so we will get:
[tex]\int\limits^1_{-1} {11 - x^2 - (-25 x + 165)} \, dx[/tex]
Simplify that to get:
[tex]\int\limits^1_{-1} {(- x^2 + 25 x - 154)} \, dx[/tex]
We will get the area:
area = [ (1/3)*( - (1)^3 - (-1)^3) - 154*(1 - (-1))
area = -308.6
A negative area means that the first function is mostly below the second one.
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When doing 2 proportion testing, you must check the Success/Failure Condition. Which of the following statements is true?
I. If both samples pass the success part but do not pass the failure part, it is a violation but does not need to be discussed in the conclusion
II. If one sample passes both parts but the other does not pass either part, it is a violation that needs to be discussed in the conclusion
III. If one sample passes both parts but the other only passes the success part, it is not a violation
IV. If both samples do not pass the success part but pass the failure part, it is a violation that must be discussed in the conclusion
a. II and III
b. I and IV
c. II and IV
The correct statement is: c. II and IV for two proportion testing.
In two proportion testing, the success/failure condition refers to the number of successes and failures in each sample. The condition states that both samples should have a sufficient number of successes and failures for the test to be valid.
II. If one sample passes both parts (has a sufficient number of successes and failures) but the other does not pass either part, it is a violation that needs to be discussed in the conclusion. This is because the sample that does not meet the success/failure condition may affect the validity and reliability of the test results.
IV. If both samples do not pass the success part (do not have a sufficient number of successes) but pass the failure part (have a sufficient number of failures), it is a violation that must be discussed in the conclusion. This violation indicates that the test may not be appropriate for analyzing the proportions in the given samples.
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Determine the volume generated of the area bounded by y=√x and y= ½ x rotated around the y-axis.
a. (64/5)π
b. (8/15)π
c. (128/25)π
d. (64/15)
To determine the volume generated by rotating the area bounded by the curves y = √x and y = ½x around the y-axis, we can use the method of cylindrical shells. By setting up the integral and evaluating it, we find that the volume is equal to (64/15)π.
To find the volume, we use the method of cylindrical shells, which involves integrating the circumference of the shells multiplied by their heights. In this case, the height of each shell is the difference between the y-values of the two curves: (√x - ½x).
We integrate with respect to x from the lower bound to the upper bound, which are the x-values where the two curves intersect: x = 0 and x = 4.
Setting up the integral and evaluating it, we find that the volume is equal to ∫(0 to 4) 2πx(√x - ½x) dx. This simplifies to (64/15)π, which is the final answer.
Therefore, the volume generated by rotating the area bounded by the curves y = √x and y = ½x around the y-axis is (64/15)π.
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If A and B are square matrices of order 3 and 2A^-1B = B - 4I,
show that A - 2I is invertible.
Given that the two matrices A and B are square matrices of order 3 and 2 respectively and, 2A⁻¹B = B - 4I. To show that A - 2I is invertible, we need to prove that det(A - 2I) ≠ 0.The equation given can be written as:2A⁻¹B = B - 4I2A⁻¹B + 4I = B2(A⁻¹B + 2I) = B
Here, B can be replaced by 2(A⁻¹B + 2I) which gives:B = 2(A⁻¹B + 2I)Now, the equation can be written as:A⁻¹B = ½(B - 4I)Now, we have two matrices, A and B, where A is a square matrix of order 3 and B is a square matrix of order 2.Given, 2A⁻¹B = B - 4I2(A⁻¹B) + 4I = BSubstituting ½(B - 4I) for A⁻¹B,
we get:2 * ½(B - 4I)A = ½(B - 4I)A = ¼(B - 4I)We know that A is a square matrix of order 3 and A - 2I is invertible, i.e. (A - 2I)⁻¹ exists. Let's assume that det(A - 2I) = 0, which means (A - 2I)⁻¹ does not exist.Therefore, det(A - 2I) ≠ 0 and (A - 2I)⁻¹ exists. So, A - 2I is invertible and the proof is complete.
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Given matrices A and B are square matrices of orders 3 and 2 respectively and 2A^−1B = B - 4I, we have to show that A - 2I is invertible.
Now, if (2A^−1 - I) is invertible, then we can write it as(2A^−1 - I)^-1 = 1/2 A(B)^-1If we multiply both sides of the equation with B, we get: B (2A^−1 - I) (1/2 A(B)^-1) = -2I(B)^-1By distributive property, it becomes:
B [(2A^-1 × 1/2A(B)^-1) - (I × 1/2A(B)^-1)] = -2I(B)^-1Let us simplify[tex]2A^-1 × 1/2A(B)^-1 = BB(B)^-1 =[/tex] I, so the equation becomes:
B (I - 1/2(B)^-1) = -2I(B)^-1Or, B [I - 1/2(B)^-1] = -2I(B)^-1Thus, (I - 1/2(B)^-1) is invertible. Thus, the matrices 2A^−1 - I and I - 1/2(B)^-1 are invertible.
As the product of two invertible matrices is also invertible, the matrix B (2A^−1 - I) (1/2 A(B)^-1) is invertible.
Now, A - 2I = (1/2)A [2A^−1 × B - 2I]Thus, we get:
A - 2I = (1/2)A [B (2A^−1 - I) (1/2 A(B)^-1) - 2I]Now, we know that the product of invertible matrices is invertible.
So,[tex]B (2A^−1 - I) (1/2 A(B)^-1[/tex]) is invertible. And so, [tex](B (2A^−1 - I) (1/2 A(B)^-1) - 2I)[/tex]is also invertible. Finally, (1/2)A [B (2A^−1 - I) (1/2 A(B)^-1) - 2I] is invertible.So, A - 2I is invertible. Hence, this is the required proof and we have shown that A - 2I is invertible.
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question 8 and 9
8- f(t)=e³¹ cos2t 9- f(t)=3+e²2¹-sinh 5t 10- f(t)=ty'.
The integral of f(t) = e³¹ cos²t is (1/2)e³¹t + (1/4)e³¹sin(2t) + C. The integration involves using the power-reducing formula for cosine squared and the substitution method.
The integral of f(t) = e³¹ cos²t is (1/2)e³¹t + (1/4)e³¹sin(2t) + C. To know more about the integration of exponential functions and trigonometric functions, refer here: [link to a reliable mathematical resource].
To integrate f(t) = e³¹ cos²t, we can use the power-reducing formula for cosine squared:
cos²t = (1/2)(1 + cos(2t))
Now, we can rewrite the integral as:
∫ e³¹ cos²t dt = ∫ e³¹ (1/2)(1 + cos(2t)) dt
Distribute e³¹ throughout the integral:
= (1/2) ∫ e³¹ dt + (1/2) ∫ e³¹ cos(2t) dt
Integrating e³¹ with respect to t gives:
= (1/2) e³¹t + (1/2) ∫ e³¹ cos(2t) dt
To integrate ∫ e³¹ cos(2t) dt, we can use the substitution method. Let u = 2t, then du = 2 dt:
= (1/2) e³¹t + (1/4) ∫ e³¹ cos(u) du
Integrating e³¹ cos(u) du gives:
= (1/2) e³¹t + (1/4) e³¹sin(u) + C
Substituting back u = 2t:
= (1/2) e³¹t + (1/4) e³¹sin(2t) + C
Therefore, the integral of f(t) = e³¹ cos²t is (1/2)e³¹t + (1/4)e³¹sin(2t) + C.
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Compute the line integral SCF.dr for F(x, y, z) = eyi + (xe + e)j + ye’k along the line segment connecting (0,2,0) to (4,0,3). = 6 none of these -5
The line integral SCF.dr for [tex]F(x, y, z) = eyi + (xe + e)j + yek[/tex] along the line segment connecting (0, 2, 0) to (4, 0, 3) is -5. Therefore, the correct answer is (D) -5.
To calculate line integral, we must use the following formula:
`∫CF.dr = ∫a(b) F(r(t)).r'(t)
dt where r(t) is the position vector given by:
[tex]r(t) = x(t)i + y(t)j + z(t)k[/tex].
We have the initial and final point of the line segment as(0, 2, 0) and (4, 0, 3) respectively.
Hence, the position vector equation is:
[tex]r(t) = (4t/4)i + (2 - 2t/4)j + (3t/4)k[/tex]
= ti + (2 - t/2)j + (3t/4)k
We obtain the denominator 4 by finding the maximum difference between the coordinates, i.e.,
Substituting the equation into the formula:
∫CF.dr=∫a(b) F(r(t)).r'(t)
dt=∫[tex]0(1) F(ti (2 - t/2), 3t/4).(i - j/2 + 3k/4)dt[/tex]
=[tex]∫0(1) [e(2-t/2)i + (te + e)(-j/2) + (3ye') 3k/4].(i - j/2 + 3k/4)dt[/tex]
=∫[tex]0(1) [(e(2-t/2) - (te + e)/2 + 9ye'/16) dt[/tex]
=∫[tex]0(1) [(2e - e(1/2)t - te/2 + 9yt/16) dt[/tex]
= (2e - (2/3)e + (1/4)e + (9/32)) - 2e
= -5
Therefore, the answer is (D) `-5`
Therefore, the line integral SCF.dr for[tex]F(x, y, z) = eyi + (xe + e)j + yek[/tex]along the line segment connecting (0,2,0) to (4,0,3) is -5.
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The cost of producing 6000 face masks is $25,600 and the cost of producing 6500 face masks is $25.775. Use this information to create a function C (a) that represents the cost in dollars a company spends to manufacture x thousand face masks during a month. The linear equation is: C (x) = ____________
The vertical intercept for this graph is at the point ____________ (type a point) and represents a cost of $ ___________when a quantity of _________face masks are produced. The rate of change for C(a) is __________and means the cost is Based on this model, C(11) = ________ which means that when a quantity of ____________ face marks are produced, there is a cost of $ _________
Solving C (a)= 90, 700 shows x = ___________ which represents that for a cost of $. you can produce _____ face masks The appropriate domain of this function is ________ (interval notation- use INF for infinity if needed).
Write the hypothesis for the following cases:
1- A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles.
2- A random survey of 75 death row inmates revealed that the mean length of time on death row is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis test to determine if the population mean time on death row could likely be 15 years, what would the null and alternative hypotheses be?
The null and alternative hypothesis are significant.
1) Hypothesis is a proposed explanation made on the basis of limited evidence as a starting point for further investigation. For the given case, the hypothesis can be stated as:
Null Hypothesis (H0): The average lifespan of the deluxe tire is greater than or equal to 50,000 miles.
Alternative Hypothesis (Ha): The average lifespan of the deluxe tire is less than 50,000 miles.
2) The null hypothesis states that there is no statistically significant difference between the two groups being tested.
It is often denoted by H0.
The alternative hypothesis is often denoted by Ha and states that there is a statistically significant difference between the two groups being tested.In this case, the null and alternative hypotheses would be:Null Hypothesis (H0):
The population mean time on death row is 15 years.
Alternative Hypothesis (Ha): The population mean time on death row is not 15 years.
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Find an equation of the tangent plane to the given surface at the specified point. z = 3(x - l)^2 + 2(y + 3)^2 + 7, (4, 1, 66) Recall that the equation of the plane tangent to z = f(x, y) at a point (a, b, c) is given by z = c c = f_x (a b) (x - a) + f_y (a b) (y - b b). For z = f(x, y) = 3(x - 1)^2 + 2(y + 3)^2 + 7, we have f_x(x, y) = and f_y(x, y) =
The equation of the tangent plane to the given surface at the specified point is 18x + 16y - 34.
Given: z = 3(x - 1)² + 2(y + 3)² + 7
We have to find the equation of the tangent plane to the given surface at the specified point.
We have a formula to find the equation of the plane tangent to z = f(x, y) at a point (a, b, c) as shown below:
z = c + [tex]f_x[/tex](a, b) (x - a) + [tex]f_y[/tex] (a, b) (y - b)
Here, we need to find [tex]f_x[/tex] (a, b) and [tex]f_y[/tex] (a, b).
Differentiating z = 3(x - 1)² + 2(y + 3)² + 7 partially with respect to x, we get:
∂z/∂x = 6(x - 1)
Differentiating z = 3(x - 1)² + 2(y + 3)² + 7 partially with respect to y, we get:
∂z/∂y = 4(y + 3)
Therefore, at point (4, 1), we have a = 4,
b = 1,
c = 66,
[tex]f_x[/tex] (a, b) = ∂z/∂x
= 6(4 - 1)
= 18
and [tex]f_y[/tex] (a, b) = ∂z/∂y
= 4(1 + 3)
= 16
Now substituting these values in the plane equation, we get:
z = 66 + 18(x - 4) + 16(y - 1)
Simplifying the above equation, we get the equation of the tangent plane as shown below:
z = 18x + 16y - 34
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Question 18 1 points Save An Which of the following statement is correct about the brands and bound algorithm derived in the lectures to solve the max cliquer problem The algorithm is better than bruteforce enumeration because its complexity is subexponential o White the algorithm is not better than tre force enameration tas both have exponential comploty, it can more often as in general do not require the explide construction of all the feasible solutions to the problem The algorithms morient than the force enumeration under no circumstances will construct the set of fantiles
The correct statement about the brands and bound algorithm derived in the lectures to solve the max cliquer problem is that it is not better than brute force enumeration in terms of worst-case time complexity, as both have exponential complexity.
However, the algorithm is more efficient than brute force enumeration in practice as it does not require the explicit construction of all feasible solutions to the problem. The brands and bound algorithm is a heuristic approach that tries to eliminate parts of the search space that are guaranteed not to contain the optimal solution. This means that the algorithm can often find the solution much faster than brute force enumeration. Additionally, the algorithm does not construct the set of cliques/families under any circumstances, which reduces the memory usage of the algorithm.
Overall, while the brands and bound algorithm may not be the most efficient algorithm for solving the max cliquer problem in theory, it is a practical and useful approach for solving the problem in real-world scenarios.
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find the values of x for which the series converges. (enter your answer using interval notation.) [infinity] (−9)nxn n = 1
The given series is `[infinity] (−9)nxn n = 1`. We need to find the values of x for which the series converges. (enter your answer using interval notation.)
To solve the problem, we will use the ratio test to determine the convergence of the given series.Ratio test: Suppose that `∑an` is a series such that `an≠0` for infinitely many n and the limit` L = lim(n→∞) |an+1/an|` exists. Then the series `∑an` is convergent if `L < 1` and divergent if `L > 1`. If `L = 1` or does not exist, the test is inconclusive.Now let's apply the ratio test to our series. Let's evaluate the limit: `lim(n→∞) |(-9)(n+1) x^(n+1)/(-9)nx^n|` `= lim(n→∞) |(-9) x|` `= |(-9) x|`.Thus, the series converges when `|(-9) x| < 1`.This is possible when: $$-1 < -9x < 1$$$$1/9 > x > -1/9$$Therefore, the values of x for which the given series converges are `[-1/9, 1/9]`. Hence, the answer is `[-1/9, 1/9]`.
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The given series is `[infinity] (−9) nxn n = 1`. We need to find the values of x for which the series converges.
To solve the problem, we will use the ratio test to determine the convergence of the given series. Ratio test:
Suppose that `∑an` is a series such that `an≠0` for infinitely many n and the limit` L = lim(n→∞) |an+1/an|` exists.
Then the series `∑an` is convergent if `L < 1` and divergent if `L > 1`. If `L = 1` or does not exist, the test is in conclusive.
Now let's apply the ratio test to our series. Let's evaluate the limit: `lim (n→∞) |(-9)(n+1) x^(n+1)/(-9) nxⁿ|` `
= lim(n→∞) |(-9) x|` `= |(-9) x|`.
Thus, the series converges when `|(-9) x| < 1.
This is possible when: $$-1 < -9x < 1$$$$1/9 > x > -1/9$$Therefore, the values of x for which the given series converges are `[-1/9, 1/9]`.
Hence, the answer is `[-1/9, 1/9]`.
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please answer all 3 questions thank you so much!
Find the equation of the curve passing through (1,0) if the slope is given by the following. Assume that x>0. dy 3 4 + dx y(x) = (Simplify your answer. Use integers or fractions for any numbers in the
To find the equation of the curve passing through (1,0) with the given slope
a) y = x^5 + 4x - 5
b) y = -1/(2x^2) + 2x - 3/2
c) y = -cos(x) + sin(x) + cos(1) - sin(1)
What are the equations of the curves passing through (1,0) with the given slopes?
We can integrate the slope function with respect to x.
a) For dy/dx = 3x^4 + 4, we integrate both sides with respect to x:
∫dy = ∫(3x^4 + 4)dx
Integrating, we get:
y = x^5 + 4x + C
Substituting the point (1,0), we can solve for the constant C:
0 = (1^5) + 4(1) + C
0 = 1 + 4 + C
C = -5
Therefore, the equation of the curve passing through (1,0) is:
y = x^5 + 4x - 5.
b) Similarly, for y(x) = (1/x^3) + 2, the integration gives:
y = -1/(2x^2) + 2x + C
Substituting (1,0) gives:
0 = -1/(2(1)^2) + 2(1) + C
0 = -1/2 + 2 + C
C = -3/2
So, the equation of the curve is:
y = -1/(2x^2) + 2x - 3/2.
c) Lastly, for dy/dx = sin(x) + cos(x), integrating yields:
y = -cos(x) + sin(x) + C
Using the given point (1,0):
0 = -cos(1) + sin(1) + C
C = cos(1) - sin(1)
Thus, the equation of the curve is:
y = -cos(x) + sin(x) + cos(1) - sin(1).
The constant C represents the arbitrary constant of integration, which is determined by the initial condition or the given point on the curve.
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Use the modified Euler's method to obtain an approximate solution of --21.) -1, in the interval di Osts 0.5 using ) - 0.1. Compute the error and the percentage error. Given the exact solution is given by y = (+7 Solution: For n-0: y/- % -26-1-20.1) (0) (19-1 Now x = x + (-2698 – 24 %108] - 1 - (0.180) (1° +(0.1)09) - 0,99 Table E8.12 shows the remaining calculations. Table E8.12 also shows the values obtained from the Euler's method, the modified Euler's method, the exact values, and the percentage error for the modified Euler's method Table E8.12 Euler Modified Exact Error Percentage Y. Euler ya value Error 00 1 1 1 0 0 10.1 1 0.9900 0.9901 0.0001 0.0101 20.2 0.9800 0.9614 0.9615 0.0001 0.0104 30.3 0.9416 0.9173 0,9174 0.0001 0.0109 4 0.4 0.8884 0.8620 0.8621 0.0001 0.0116 5 0.5 0.8253 0.8001 0.8000 0.0001 0.0125 In the Table E8.12. Error exact Value - value from modified Euler's method - error Percentage error exact value
The differential equation for which modified Euler's method is used to obtain an approximate solution is given by: dy/dx = -2y, y(0) = -1. The approximate solution will be computed using h = 0.1 on the interval [0, 0.5].Steps for Modified Euler's Method are:
Step 1: Find y1 using Euler's Methody 1 = y0 + hf(x0, y0)Where y0 = -1 and x0 = 0, so thatf(x, y) = -2y.Hence, y1 = -1 + 0.1(-2(-1)) = -0.8
Step 2: Find y2 using Modified Euler's Method y2 = y1 + h/2(f(x1, y1) + f(x0, y0))Where x1 = 0.1 and y1 = -0.8Therefore,f(x1, y1) = -2(-0.8) = 1.6f(x0, y0) = -2(-1) = 2Thus, y2 = -0.8 + 0.1/2(1.6 + 2) = -0.66
Step 3: Find y3 using Modified Euler's Method y3 = y2 + h/2(f(x2, y2) + f(x1, y1))Where x2 = 0.2 and y2 = -0.66Therefore,f(x2, y2) = -2(-0.66) = 1.32f(x1, y1) = -2(-0.8) = 1.6.
Thus, y3 = -0.66 + 0.1/2(1.32 + 1.6) = -0.548Step 4: Find y4 using Modified Euler's Methody4 = y3 + h/2(f(x3, y3) + f(x2, y2)).
Where x3 = 0.3 and y3 = -0.548.Therefore,f(x3, y3) = -2(-0.548) = 1.096f(x2, y2) = -2(-0.66) = 1.32Thus, y4 = -0.548 + 0.1/2(1.096 + 1.32) = -0.4448
Step 5: Find y5 using Modified Euler's Methody5 = y4 + h/2(f(x4, y4) + f(x3, y3))Where x4 = 0.4 and y4 = -0.4448
Therefore,f(x4, y4) = -2(-0.4448) = 0.8896f(x3, y3) = -2(-0.548) = 1.096.
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Use the Golden Search method to maximize the following unimodal function, ƒ(X) = −(x − 3)², 2 ≤ x ≤ 4 with A = 0.05.
We will use the Golden Section Search method to maximize the unimodal function ƒ(x) = -(x - 3)² within the interval 2 ≤ x ≤ 4, with an accuracy level of A = 0.05.
The Golden Section Search is an optimization algorithm that narrows down the search interval iteratively by dividing it in a specific ratio based on the golden ratio. In each iteration, we evaluate the function at two points within the interval and compare the function values to determine the new search interval.
To apply the Golden Section Search, we start with the initial interval [a, b] = [2, 4]. The interval is divided into two subintervals based on the golden ratio, giving us two points x₁ and x₂. We evaluate the function at these points and compare the function values to determine the new search interval.
In the first iteration, we evaluate ƒ(x₁) and ƒ(x₂) and compare the values. Since we want to maximize the function, if ƒ(x₁) > ƒ(x₂), we update the search interval to [a, x₂], otherwise, we update it to [x₁, b]. We continue this process iteratively, narrowing down the interval until we reach the desired accuracy level.
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A mean weight of 500 sample cars found (1000 + B) Kg. Can it be reasonably regarded as a sample from a large population of cars with mean weight 1500 Kg and standard deviation 130 Kg?
Test at 5% level of significance. (20 Marks) B = 022
The data suggests that the sample of 500 cars does not come from a population with a mean weight of 1500 Kg at a 5% level of significance.
To determine if the sample of 500 cars can be reasonably regarded as a sample from a population with a mean weight of 1500 Kg and a standard deviation of 130 Kg, we can perform a hypothesis test.
Let's set up the null and alternative hypotheses:
Null hypothesis (H0): The sample is from a population with a mean weight of 1500 Kg.
Alternative hypothesis (Ha): The sample is not from a population with a mean weight of 1500 Kg.
We can conduct a one-sample t-test to test this hypothesis. The test statistic is calculated as:
t = ([tex]\bar X[/tex] - μ) / (s / √n)
Where:
[tex]\bar X[/tex] is the sample mean weight (1000 + B)
μ is the population mean weight (1500)
s is the sample standard deviation (unknown)
n is the sample size (500)
We are given that B = 022, so the sample mean weight can be calculated as:
[tex]\bar X[/tex] = 1000 + B = 1000 + 0.022 = 1000.022 Kg
Since the sample standard deviation is unknown, we cannot directly calculate the test statistic. However, if the sample size is sufficiently large (usually considered when n > 30), we can assume that the sample standard deviation is a good estimate of the population standard deviation.
Given that we have a large sample size of 500, we can proceed with the assumption that the sample standard deviation is a good estimate of the population standard deviation (130 Kg).
Next, we calculate the t-value using the formula above and the given values:
t = (1000.022 - 1500) / (130 / √500)
Using a statistical calculator or software, we can find the critical t-value at a 5% level of significance with 499 degrees of freedom (500 - 1). The critical t-value for a one-tailed test is approximately 1.646.
If the calculated t-value is greater than the critical t-value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Calculate the t-value:
t = (1000.022 - 1500) / (130 / √500) ≈ -31.3
Since the calculated t-value (-31.3) is much smaller than the critical t-value (1.646), we reject the null hypothesis. Therefore, the sample cannot be reasonably regarded as a sample from a population with a mean weight of 1500 Kg.
In conclusion, the data suggests that the sample of 500 cars does not come from a population with a mean weight of 1500 Kg at a 5% level of significance.
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The average cost in terms of quantity is given as C(q) =q²-3q+100, the margina rofit is given as MP(q) = 3q-1. Find the revenue. (Hint: C(q) = C(q) /q, R(0) = 0)
The average cost in terms of quantity is given as C(q) =q²-3q+100, and the marginal profit is given as MP(q) = 3q-1. The revenue is given by R(q) = [4q² - 3q + 100]/q.
The average cost in terms of quantity is C(q) = q² - 3q + 100 and the marginal profit is MP(q) = 3q - 1. We have to identify the revenue. In order to identify the revenue, we have to use the relation among revenue, cost, and profit which is Revenue = Cost + Profitor, R(q) = C(q) + P(q)
Now, we have to calculate the Revenue, therefore we first need to identify the Cost and Profit. Cost is,
C(q) = q² - 3q + 100
For calculating profit, we use the relation: MP(q) = R'(q) = P(q)
Where MP(q) is the marginal profit and P(q) is the profit. R'(q) = P(q) = 3q - 1.
Putting this value in relation to Cost, we get
C(q) = C(q)/qR (q) = C(q) + P(q)
R(q) = [q² - 3q + 100]/q + [3q - 1]
Now, we simplify the above expression as follows: R(q) = [(q² - 3q + 100) + (3q² - q)]/qR(q) = [4q² - 3q + 100]/q
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$800 is invested at a rate of 4% and is compounded monthly. Find the balance after 10 years.
The balance after 10 years would be approximately $1,190.96.
To calculate the balance after 10 years of investing $800 at a rate of 4% compounded monthly, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final balance
P = the principal amount (initial investment)
r = the annual interest rate (as a decimal)
n = the number of times the interest is compounded per year
t = the number of years
In this case, we have:
P = $800
r = 4% = 0.04 (as a decimal)
n = 12 (compounded monthly)
t = 10 years
Plugging the values into the formula, we have:
A = 800(1 + 0.04/12)^(12 × 10)
Simplifying the calculation inside the parentheses:
A = 800(1 + 0.003333)^120
Using a calculator, we can evaluate (1 + 0.003333)^120 ≈ 1.4887.
A = 800 × 1.4887 ≈ $1,190.96
Therefore, the balance after 10 years would be approximately $1,190.96.
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find u, v , u , v , and d(u, v) for the given inner product defined on rn. u = (1, 0, 2, −1), v = (0, 2, −1, 1), u, v = u · v
[tex]u = (1, 0, 2, −1), v = (0, 2, −1, 1), u, v = -1[/tex] and d(u, v) = 3√2, which are the values of u, v, u, v and d(u, v)..
Given the inner product defined on Rn is given by;
u = (1, 0, 2, −1), v = (0, 2, −1, 1), u, v = u · v
To find the values of u, v, u, v and d(u, v) we use the following;
[tex]u = (u1, u2, u3, ...., un) v = (v1, v2, v3, ...., vn)d(u, v) = √⟨u − v, u − v⟩[/tex]
We can determine u and v as follows;
u = (1, 0, 2, −1), v = (0, 2, −1, 1)u1 = 1, u2 = 0, u3 = 2, u4 = -1v1 = 0, v2 = 2, v3 = -1, v4 = 1
Then u.
v is given by;
[tex]u . v = u1v1 + u2v2 + u3v3 + u4v4= (1)(0) + (0)(2) + (2)(-1) + (-1)(1)= -1[/tex]
Now we can find d(u, v) as follows;
[tex]d(u, v) = √⟨u − v, u − v⟩= √⟨(1, 0, 2, −1) - (0, 2, −1, 1), (1, 0, 2, −1) - (0, 2, −1, 1)⟩[/tex]
= [tex]√⟨(1, -2, 3, -2), (1, -2, 3, -2)⟩[/tex]
= [tex]√(1^2 + (-2)^2 + 3^2 + (-2)^2)[/tex]
= [tex]√(1 + 4 + 9 + 4)= √18 = 3√2[/tex]
Therefore;
u = (1, 0, 2, −1), v = (0, 2, −1, 1), u, v = -1 and d(u, v) = 3√2, which are the values of u, v, u, v and d(u, v)..
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In a class of 32 students, there are 14 students that play on a sports team and 12 students that play in one of the school bands. There are 8 students that do not play a sport or play in a band. Some play on a team and play in one of the bands. What is the probability that a student chosen at random will play on a sports team or play in one of the school bands?
The probability that a student chosen at random will play on a sports team or play in one of the school bands is 75%. The number of students who play both in a sports team and in one of the school bands is 24 students.
There are two ways to find out the number of students who play both in a sports team and in one of the school bands:1.
We can use a Venn diagram or2. Use the formula, n(A ∩ B) = n(A) + n(B) - n(A ∪ B)
Let us use the Venn diagram approach to find out the number of students who play both in a sports team and in one of the school bands.
A Venn diagram is a graphical representation of the relationships between sets.
The sample space, which is the set of all possible outcomes, is represented by a rectangle.
Each set is represented by a circle or an oval. The overlapping region represents the intersection of two or more sets.
The non-overlapping regions represent the sets themselves and their complements (the elements that do not belong to the set).
Here,14 students play on a sports team,12 students play in one of the school bands, and8 students do not play a sport or play in a band.
To find n(A ∩ B), we can use the formula,n(A ∩ B) = n(A) + n(B) - n(A ∪ B)
Here, n(A ∪ B) represents the total number of students who play on a sports team or play in one of the school bands.n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
So, n(A ∩ B) = n(A) + n(B) - n(A ∪ B)= 14 + 12 - (32 - 8)= 24 students.
Therefore, the number of students who play both in a sports team and in one of the school bands is 24 students.
Total number of students who play in a sports team or play in one of the school bands = n(A ∪ B)= n(A) + n(B) - n(A ∩ B)= 14 + 12 - 24= 26 students
Hence, the probability that a student chosen at random will play on a sports team or play in one of the school bands is P(A)
= (Number of favorable outcomes) / (Total number of outcomes)
= (26 + 24) / 32= 50 / 64= 75%.
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o make a specific prediction for an individual's score on a given variable, when we know the individual's score on two or more correlated variables, we would use what statistical technique? a. Linear regression b. Multiple correlation coefficient c. Pearson's r correlation coefficient d. Multiple regression
When we want to make a specific prediction for an individual's score on a given variable, when we know the individual's score on two or more correlated variables, we would use the statistical technique known as Multiple Regression.
Multiple Regression is a statistical technique used to assess the relationship between a dependent variable and one or more independent variables. It is used when we need to understand how the value of the dependent variable changes with changes in one or more independent variables. Multiple regression is used when we want to predict a continuous dependent variable from a number of independent variables. In multiple regression, we are interested in the regression equation that uses one or more independent variables to predict a dependent variable. The conclusion of a multiple regression analysis provides information about the relationship between the dependent variable and the independent variables. It tells us whether the relationship is statistically significant, the strength of the relationship, and the direction of the relationship.
Thus, the correct option is (d) Multiple Regression.
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Use Shell method to find the volume of the solid formed by revolving the region bounded by the graph of y=x³+x+l, y = 1 and X=1 about the line X = 2₁"
To calculate the flux of the vector field F = (x/e)i + (z-e)j - xyk across the surface S, which is the ellipsoid x²/25 + y²/5 + z²/9 = 1, we can use the divergence theorem.
The divergence theorem states that the flux of a vector field across a closed surface is equal to the triple integral of the divergence of the vector field over the volume enclosed by the surface.
First, let's calculate the divergence of F:
div(F) = (∂/∂x)(x/e) + (∂/∂y)(z-e) + (∂/∂z)(-xy)
= 1/e + 0 + (-x)
= 1/e - x
To calculate the surface integral of the vector field F = (x/e) I + (z-e)j - xyk across the surface S, which is the ellipsoid x²/25 + y²/5 + z²/9 = 1, we can set up the surface integral ∬S F · dS.
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Let f(x) = √2x - 10 and the virtual line joining the origin (0, 0) to a point Q moving on the curve of the function f. The curve and the line are shown below. a) Determine the coordinates of point Q that would maximize the viewing angle theta (0) of an observer whose eye, located at the origin, follows the displacement of the point Q along the curve. Note that tan(0) = b) Determine this maximum angle (in degrees)
To determine the coordinates of point Q that would maximize the viewing angle θ(0) and find the maximum angle in degrees, we need to find the maximum value of the tangent function.
Given that f(x) = √(2x) - 10, we want to find the maximum value of tan(θ(0)).
The tangent function is defined as tan(θ) = opposite/adjacent, which in this case is y/x.
Let's find the equation of the line connecting the origin (0, 0) to point Q on the curve.
The equation of the line is y = mx, where m is the slope of the line.
The slope, m, is given by m = (f(x) - 0)/(x - 0) = f(x)/x.
Substituting f(x) = √(2x) - 10, we have m = (√(2x) - 10)/x.
Now, let's substitute y = mx into the equation of the curve:
√(2x) - 10 = (√(2x) - 10)/x * x.
Simplifying, we have:
√(2x) - 10 = (√(2x) - 10).
Both sides of the equation are equal, indicating that any point on the curve satisfies this equation.
To maximize the viewing angle θ(0), we need to find the point Q on the curve where the tangent function tan(θ(0)) is maximized.
The tangent function is maximized when the slope of the line connecting the origin to point Q is maximized. This occurs when the line is tangent to the curve.
To find the point Q where the line is tangent to the curve, we need to find the maximum value of the slope (√(2x) - 10)/x.
Taking the derivative of the slope with respect to x and setting it equal to zero to find the critical points:
d/dx [(√(2x) - 10)/x] = 0.
Using the quotient rule for differentiation, we get:
[(1/2√(2x))x - (√(2x) - 10)]/x^2 = 0.
Simplifying, we have:
(1/2√(2x))x - (√(2x) - 10) = 0.
Solving for x, we find:
x = 20.
Now, we substitute x = 20 into the equation of the line to find the y-coordinate of point Q:
y = (√(2x) - 10) = (√(2*20) - 10) = 0.
Therefore, the coordinates of point Q that maximize the viewing angle θ(0) are (20, 0).
Now, to determine the maximum angle θ(0) in degrees, we can calculate it using the arctan function:
θ(0) = arctan(m) = arctan((√(2x) - 10)/x) = arctan((√(2*20) - 10)/20) ≈ 43.60 degrees.
Therefore, the maximum angle θ(0) is approximately 43.60 degrees.
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pls clear hand writing
a) The sum of the first n terms of the progression 36,34,32, ...is 0. Find n and the tenth (4 marks) term.
n = 37, and tenth term = 18
Given progression,
36, 34, 32, ...
The sum of the first n terms is 0
First term(a1) = 36
The common difference (d)= 34-36 = -2,
The formula of the sum of the first n term is,
[tex]Sn = \frac{n}{2} [2a_{1} + (n - 1)d][/tex]
substitue the values Sn= 0, a1= 36, d= -2 in the above equation to find n
[tex]0[/tex]= [tex]\frac{n}{2} [2(36) + (n-1) (-2)][/tex]
[tex]0 = \frac{n}{2}[72- 2n+ 2][/tex]
[tex]0 = \frac{n}{2}[74 - 2n][/tex]
[tex]74 - 2n = 0[/tex]
[tex]2n = 74[/tex]
[tex]n = \frac{74}{2}[/tex]
[tex]n = 37[/tex]
n = 37
The formula for finding the nth term(10th term):
[tex]a_{n} = a1 + (n - 1)d[/tex]
n = 10, a1 = 36, d = -2
[tex]a_{10} = 36 + (10-1)(-2)[/tex]
[tex]a_{10} = 36 + 9(-2)[/tex]
[tex]a_{10} = 36 - 18[/tex]
[tex]a_{10} = 18[/tex]
[tex]a_{10}[/tex] = 18
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Find the average rate of change of f(x) = 4x² - 5 on the interval [3, t). Your answer will be an expression involving t .
Given, the function is f(x) = 4x² - 5 and the interval is [3, t).
We have to find the average rate of change of f(x) on the interval [3, t).
The average rate of change of f(x) on the interval [a, b] is given by:
(f(b) - f(a))/(b-a)
To find the average rate of change of f(x) on the interval [3, t), we have to put a = 3 and b = t in the above formula.
Average rate of change = (f(t) - f(3))/(t-3)
Average rate of change = (4t² - 5 - 4(3)² + 5)/(t-3)
Average rate of change = (4t² - 32)/(t-3)
Therefore, the expression involving t that represents the average rate of change of f(x) on the interval [3, t) is:
(4t² - 32)/(t-3)
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