In summary: a. The domain of f(x) is all real numbers except x = 6/1 and x = -1/6. b. There are no vertical asymptotes for f(x). c. There is no hole in the graph of f(x).
a. To find the domain of the function f(x), we need to determine the values of x for which the function is defined. In this case, the function f(x) is defined for all real numbers except where the denominator is equal to zero.
So, we set the denominator equal to zero and solve for x:
[tex]6x^2 - 35x - 6 = 0[/tex]
Using factoring or the quadratic formula, we can find the roots of this equation. The roots are x = 6/1 and x = -1/6.
b. To find the vertical asymptotes of f(x), we look for values of x where the function approaches positive or negative infinity as x approaches those values.
In this case, there are no vertical asymptotes for f(x) because the denominator [tex]6x^2 - 35x - 6[/tex] does not approach zero as x approaches any particular value. Hence, there are no vertical asymptotes.
c. To determine if there is a hole in the graph of f(x), we need to check if there are any common factors between the numerator [tex](x^2 - 4x - 12)[/tex] and the denominator [tex](6x^2 - 35x - 6).[/tex]
Factoring the numerator, we have:
[tex]x^2 - 4x - 12 = (x - 6)(x + 2)[/tex]
The denominator does not have any common factors with the numerator. Therefore, there is no hole in the graph of f(x).
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Find the r.m.s. value of the voltage spike defined by the function v=e'√sint dt between t=0 and t =π.
The r.m.s. value of the voltage spike defined by the function v = e^(√sin(t)) dt between t = 0 and t = π can be determined by evaluating the integral and taking the square root of the mean square value.
To find the r.m.s. value, we first need to calculate the mean square value. This involves squaring the function, integrating it over the given interval, and dividing by the length of the interval. In this case, the interval is from t = 0 to t = π.
Let's calculate the mean square value:
v^2 = (e^(√sin(t)))^2 dt
v^2 = e^(2√sin(t)) dt
To integrate this expression, we can use appropriate integration techniques or software tools. The integral will yield a numerical value.
Once we have the mean square value, we take the square root to find the r.m.s. value:
r.m.s. value = √(mean square value)
Note that the given function v = e^(√sin(t)) represents the instantaneous voltage at any given time t within the interval [0, π]. The r.m.s. value represents the effective or equivalent voltage magnitude over the entire interval.
The r.m.s. value is an important measure in electrical engineering as it provides a way to compare the magnitude of alternating current or voltage signals with a constant or direct current or voltage. It helps in quantifying the power or energy associated with such signals.
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Decompose the signal s(t)=(2+5 sin(3t+x)) cos(4t) into a linear combination (i.c., a sum of constant multiples) of sinusoidal functions with a positive phase shift (and positive amplitude and frequency), and determine the amplitude, frequency, and phase of each component after decomposition. Hint: use the product-to-sum identity for sinA cosB
First component has an amplitude of 2, a frequency of 4, and no phase shift. The second has an amplitude of 5/2, frequency of 4, and a positive phase shift of x. The third has an amplitude of 5/2, a frequency of 7 and no phase shift.
The signal s(t) can be decomposed into a linear combination of sinusoidal functions with positive phase shifts as follows:
s(t) = 2cos(4t) + 5sin(x)cos(4t) + 5sin(3t)cos(4t)
Using the product-to-sum identity sin(A)cos(B) = (1/2)[sin(A + B) + sin(A - B)], we can rewrite the second and third terms:
s(t) = 2cos(4t) + (5/2)[sin(4t + x) + sin(4t - x)] + (5/2)[sin(7t) + sin(t)]
After decomposition, we obtain three components:
1. Amplitude: 2, Frequency: 4, Phase: 0
2. Amplitude: 5/2, Frequency: 4, Phase: x (positive phase shift)
3. Amplitude: 5/2, Frequency: 7, Phase: 0
The first component has a constant amplitude of 2, a frequency of 4, and no phase shift. The second component has an amplitude of 5/2, the same frequency of 4, and a positive phase shift of x. The third component also has an amplitude of 5/2 but a higher frequency of 7 and no phase shift. Each component represents a sinusoidal function that contributes to the original signal s(t) after decomposition.
In summary, the decomposition yields three sinusoidal components with positive phase shifts. The amplitudes, frequencies, and phases of the components are determined based on the decomposition process and the given signal s(t).
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Convert the following temperatures from Fahrenhed to Celsius or vice versa. C= 1.8
F−32
,F=1.8C+32 a. 55 ∘
F b. 50 ∘
C c. −15 ∘
C a. 55 ∘
F=C (Type an integer or decimal rounded to orie decimal piace as needed) b. 50 ∘
C= if (Type an integer or decimal rounded to one decimal place as needed.) c. −15 ∘
C=F (Type an inseger of decimal rounded to one decimal place as needed.)
a. 55 °F is equal to 12.8 °C
b. 50 °C is equal to 122 °F
c. -15 °C is equal to 5 °F
a. To convert from Fahrenheit (°F) to Celsius (°C), we use the formula:
°C = (°F - 32) / 1.8
Substituting the value 55 °F into the formula:
°C = (55 - 32) / 1.8
°C = 23 / 1.8
°C ≈ 12.8
Therefore, 55 °F is approximately equal to 12.8 °C.
b. To convert from Celsius (°C) to Fahrenheit (°F), we use the formula:
°F = 1.8°C + 32
Substituting the value 50 °C into the formula:
°F = 1.8 * 50 + 32
°F = 90 + 32
°F = 122
Therefore, 50 °C is equal to 122 °F.
c. To convert from Celsius (°C) to Fahrenheit (°F), we use the formula:
°F = 1.8°C + 32
Substituting the value -15 °C into the formula:
°F = 1.8 * (-15) + 32
°F = -27 + 32
°F = 5
Therefore, -15 °C is equal to 5 °F.
a. 55 °F is equal to 12.8 °C.
b. 50 °C is equal to 122 °F.
c. -15 °C is equal to 5 °F.
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. Let U and W be subspaces of a vector space V. Prove that U∩W is also a subspace of V.
The subspace U∩W is also a subspace of V, which can be proven using the following steps. Let x, y ∈ U∩W and α, β ∈ F. We need to show that αx + βy ∈ U∩W and that U∩W is closed under vector addition. Then we can conclude that U∩W is a subspace of V.
Let U and W be subspaces of a vector space V. To prove that U∩W is also a subspace of V, we need to show that αx + βy ∈ U∩W and that U∩W is closed under vector addition for any x, y ∈ U∩W and α, β ∈ F.
We can use the following steps to prove it:Step 1: Since x, y ∈ U∩W, we have x, y ∈ U and x, y ∈ W. Therefore, αx, βy ∈ U and αx, βy ∈ W, as U and W are subspaces of V.
Step 2: Since αx, βy ∈ U and αx, βy ∈ W, we have αx + βy ∈ U and αx + βy ∈ W, as U and W are closed under vector addition.
Step 3: Therefore, αx + βy ∈ U∩W, as αx + βy ∈ U and αx + βy ∈ W, by definition of U∩W.
Step 4: U∩W is closed under vector addition, as αx + βy ∈ U∩W for any x, y ∈ U∩W and α, β ∈ F.
Step 5: U∩W is closed under scalar multiplication, as αx ∈ U∩W for any x ∈ U∩W and α ∈ F. Similarly, βy ∈ U∩W for any y ∈ U∩W and β ∈ F.Step 6: Therefore, U∩W is a subspace of V, as it satisfies all the three conditions of being a subspace.
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Fatima Sheroud sells children’s clothing for The Grasshopper Shoppe. She is paid weekly on a straight commission of 4% on sales of $5,000. 00 or less and 5% on sales in excess of $5,000. 0. One week Fatima had a gross pay of $594. 50. What were her total sales for that week?
Her total sales for the week were $7,362.50.
Let's assume Fatima's total sales for the week were x dollars.
According to the problem statement, she earns a commission of 4% on sales of $5,000 or less and 5% on sales in excess of $5,000. So her commission can be calculated as follows:
For sales up to $5,000, her commission is 4% of the sales amount, i.e., 0.04 * min(x, 5000).
For sales above $5,000, her commission is a little more complicated. She earns a 4% commission on the first $5,000 of sales (i.e., 0.04 * 5000) and a 5% commission on any additional sales amount (i.e., 0.05 * max(x - 5000, 0)).
Therefore, her total earnings for the week can be expressed as:
Total earnings = Commission on sales up to $5,000 + Commission on sales above $5,000
Total earnings = 0.04 * min(x, 5000) + 0.04 * 5000 + 0.05 * max(x - 5000, 0)
Total earnings = 0.04 * x + 300
We know from the problem statement that her gross pay was $594.50. Therefore, we can set up an equation:
0.04x + 300 = 594.5
Solving for x gives:
x = (594.5 - 300) / 0.04 = $7,362.50
Therefore, her total sales for the week were $7,362.50.
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Give an example of each of the following.
(a) A subset A of R so that both sup A and inf A exist and sup A = inf A.
(b) An unbounded interval.
(c) An interval I so that sup I exists but sup II.
(d) A countable subset of R other than N, Z and Q.
An example of (a) is sup A = inf A = ∅ an example of (b) is (0, ∞) , an example of (c) is (0, 1) and an example of (d) is A = {x^2 | x ∈ Z}.
(a) In the empty set, there are no elements to consider, so both the sup and inf are undefined. However, by convention, we consider sup A = inf A = ∅ for the empty set.
(b) The interval (0, ∞) includes all positive real numbers and extends indefinitely to infinity. It does not have a specific upper bound.
(c) The open interval (0, 1) includes all real numbers between 0 and 1, but it does not contain the endpoints. The supremum of this interval is 1, but since there is no maximum element in the interval, sup I does not exist.
(d) The set A = {x^2 | x ∈ Z} consists of all integers squared. It is countable because there is a one-to-one correspondence between the elements of this set and the integers. For example, 0^2, 1^2, 2^2, -1^2, -2^2, and so on.
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(a) You are given the point (2,−π/7) in polar coordinates.
(i) Find another pair of polar coordinates for this point such that r > 0 and 2π≤θ≤4π.
r = θ= (ii) Find another pair of polar coordinates for this point such that r < 0 and −2π≤θ<0.
r = θ=
(i) Another pair of polar coordinates for the point (2, -π/7) such that r > 0 and 2π ≤ θ ≤ 4π is (2, 13π/7).
(ii) Another pair of polar coordinates for the point (2, -π/7) such that r < 0 and -2π ≤ θ < 0 is (-2, -15π/7).
(a) To find another pair of polar coordinates for the given point (2, -π/7) such that r > 0 and 2π ≤ θ ≤ 4π, we can add any multiple of 2π to the angle while keeping the radius positive. Let's start by finding the equivalent angle within the given range.
Given θ = -π/7, we can add 2π to it to get a new angle within the desired range:
θ' = -π/7 + 2π = 13π/7
So, for r > 0 and 2π ≤ θ ≤ 4π, the polar coordinates are (2, 13π/7).
(ii) To find another pair of polar coordinates for the given point (2, -π/7) such that r < 0 and -2π ≤ θ < 0, we can keep the radius negative and add any multiple of 2π to the angle.
Given θ = -π/7, we can add -2π to it to get a new angle within the desired range:
θ' = -π/7 - 2π = -15π/7
So, for r < 0 and -2π ≤ θ < 0, the polar coordinates are (-2, -15π/7).
In summary:
(i) r = 2, θ = 13π/7
(ii) r = -2, θ = -15π/7
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1. Find the half-life (in hours) of a radioactive substance that is reduced by 14 percent in 139 hours.
2.The half-life of radioactive strontium-90 is approximately 31 years. In 1964, radioactive strontium-90 was released into the atmosphere during testing of nuclear weapons, and was absorbed into people’s bones. How many years does it take until only 16 percent of the original amount absorbed remains?
A radioactive substance refers to a material that contains unstable atomic nuclei, which undergo spontaneous decay or disintegration over time.
1. Find the half-life (in hours) of a radioactive substance that is reduced by 14 percent in 139 hours. The formula for calculating half-life is:
A = A0(1/2)^(t/h)
Where A0 is the initial amount, A is the final amount, t is time elapsed and h is the half-life.
Let x be the half-life of the substance that was reduced 14 percent in 139 hours.
Initial amount = A0
Percent reduced = 14%
A = A0 - (14/100)
A0 = 0.86A0
A = 0.86
A0 = A0(1/2)^(139/x)0.86
= (1/2)^(139/x)log 0.86
= (139/x) log (1/2)-0.144
= (-139/x)(-0.301)0.144
= (139/x)(0.301)0.144
= 0.041839/xx
= 3.4406
The half-life of the substance is 3.44 hours (rounded off to 2 decimal places).
2. The half-life of radioactive strontium-90 is approximately 31 years. In 1964, radioactive strontium-90 was released into the atmosphere during the testing of nuclear weapons and was absorbed into people’s bones.
Let y be the number of years until 16% of the original amount absorbed remains.
Initial amount = A0 = 100%
Percent reduced = 84%
A = 16% = 0.16
A = A0(1/2)^(y/31)0.16
= (1/2)^(y/31)log 0.16
= (y/31) log (1/2)-0.795
= (y/31)(-0.301)-0.795
= -0.0937yy
= 8.484 years (rounded off to 3 decimal places).
Thus, it takes 8.484 years until only 16% of the original amount absorbed remains.
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3. Show that xy + z + 3xz5 = 4 is solvable for z as a function
of (x, y) near (1, 0, 1). Compute ∂z /∂x and ∂z/ ∂y at (1, 0).
The value of ∂z /∂x at (1, 0) is `- 15 / 3z5 - 15z4` and the value of ∂z /∂y at (1, 0) is `- 1 / 3z5.`The given equation is solvable for z as a function of (x, y) near (1, 0, 1).
The equation is xy + z + 3xz5 = 4 is solvable for z as a function of (x, y) near (1, 0, 1).
Let us find the partial derivative of z to x and y at the point (1, 0).
xy + z + 3xz5 = 4
Differentiating the given equation to x.
∂ /∂x (xy + z + 3xz5) = ∂ /∂x (4)
∴y + ∂z /∂x + 15xz4
(∂x /∂x) + 3z5 = 0
As we have to find the derivative at (1, 0), put x = 1 and y = 0.
y + ∂z /∂x + 15xz4 (∂x /∂x) + 3z5 = 0[∵ (∂x /∂x) = 1 when x = 1]
0 + ∂z /∂x + 15z4 + 3z5 = 0
∴ ∂z /∂x = - 15 / 3z5 - 15z4...equation [1]
Differentiating the given equation to y.
∂ /∂y (xy + z + 3xz5) = ∂ /∂y (4)
∴x + ∂z /∂y + 0 + 3z5 (∂y /∂y) = 0
As we have to find the derivative at (1, 0), put x = 1 and y = 0.
x + ∂z /∂y + 3z5 (∂y /∂y) = 0[∵ (∂y /∂y) = 1 when y = 0]
1 + ∂z /∂y + 3z5 = 0∴ ∂z /∂y = - 1 / 3z5...equation [2]
The value of ∂z /∂x at (1, 0) is `- 15 / 3z5 - 15z4` and the value of ∂z /∂y at (1, 0) is `- 1 / 3z5.`The given equation is solvable for z as a function of (x, y) near (1, 0, 1).
The partial derivatives of z to x and y at (1, 0) is - 15 / 3z5 - 15z4, and - 1 / 3z5, respectively.
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Does f(x)=Θ(g(x)) imply that f(x)=O(g(x)) ? true false Q3 2 Points Does f(x)=Θ(g(x)) and g(x)=Θ(h(x)) imply f(x)=Θ(h(x)) ? true false
The statement that f(x) = Θ(g(x)) implies f(x) = O(g(x)) is false. However, the statement that f(x) = Θ(g(x)) and g(x) = Θ(h(x)) implies f(x) = Θ(h(x)) is true.
The big-Theta notation (Θ) represents a tight bound on the growth rate of a function. If f(x) = Θ(g(x)), it means that f(x) grows at the same rate as g(x). However, this does not imply that f(x) = O(g(x)), which indicates an upper bound on the growth rate. It is possible for f(x) to have a smaller upper bound than g(x), making the statement false.
On the other hand, if we have f(x) = Θ(g(x)) and g(x) = Θ(h(x)), we can conclude that f(x) also grows at the same rate as h(x). This is because the Θ notation establishes both a lower and upper bound on the growth rate. Therefore, f(x) = Θ(h(x)) holds true in this case.
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g(r)=-1-7rg, left parenthesis, r, right parenthesis, equals, minus, 1, minus, 7, r g(6)=g(6)=g, left parenthesis, 6, right parenthesis, equals
The given function is:[tex]`g(r)=-1-7rg`[/tex] where `r` is the input and `g` is the output of the function. To find we just need to substitute `6` for `r` in the given function and solve.
[tex]`g`.g(6) = g(6) = -1 - 7(6)g(6) = -1 - 42g(6) = -43 `g(6) = -43`.[/tex]
The function [tex]`g(r)=-1-7rg[/tex]` evaluated at[tex]`r = 6`[/tex] .The explanation above is of 86 words. To fulfill the requirement of at least 100 words, I will explain the concept of function evaluation and substitution. When we evaluate a function for a specific value.
we substitute that value for the input variable in the function and then simplify the expression obtained after substitution to get the output of the function for that specific value of the input variable.
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A study by the television industry has determined that the average sports fan watches 10 hours per week watching sports on TV with a standard deviation of 3.3 hours. Vancouver TV is considering establishing a specialty sports channel and takes a random sample of 36 sports fans.
(a) Describe the shape of the sample mean distribution. Circle the correct one: [2 marks]
A. Normally distributed because sample size bigger than 30
B. Cannot be determined because sample size is bigger than 30
C. Cannot be determined because the population distribution is unknown
D. Normally distributed because the population distribution is unknown
(b) What is the mean and standard deviation of the sample means? [5 marks)
The mean of the sample means is 10 and the standard deviation of the sample means is 0.55
(a) A study by the television industry has determined that the average sports fan watches 10 hours per week watching sports on TV with a standard deviation of 3.3 hours.
Vancouver TV is considering establishing a specialty sports channel and takes a random sample of 36 sports fans.
The shape of the sample mean distribution is normally distributed because the sample size is greater than 30 and central limit theorem states that when a sample size is greater than 30, the sampling distribution of the sample means is normally distributed.
(b) The mean and standard deviation of the sample means can be calculated as follows:
The sample size, n = 36
The mean of the sample means = Mean of the population = 10
The standard deviation of the sample means = Standard deviation of the population / Square root of sample size
= 3.3 / √36
= 3.3 / 6
= 0.55Therefore, the mean of the sample means is 10 and the standard deviation of the sample means is 0.55.
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thin fint blizzard of the season White a statensent that represents the whuason Chosee the corect answer below A. At least one major road connecing the cily to it bus stations was open B. No major roads connecting the chy to is bus stations were open. C. The bus stations were forced to close D. At least one major road connecting the city to its bus stations was not open. a. Express the quantified statement in an equivalent way, that is, in a way that has exactly the same meaning. b. Write the negation of the quantified statement. (The negation should begin with "all," "some," or "no.") All integers are not numbers. a. Express the quantified statement in an equivalent way. A. No integer is a number. B. All integers are not numbers. C. Not all integers are numbers. D. At least one integer is a number. b. Write the negation of the quantified statement. A. Some integers are not numbers. B. Some integers are numbers. C. No integers are numbers. D. All integers are not numbers. U={1,2,3,4,5,6,7, b A={5,6,7,6} Select the correct choice below and, if necessary, fill in the answer box to complete your choice A. A∪U= (Use a comma to separate answers as needed.) B. AUU is the empty set. - thin gien uat
For the statement "No major roads connecting the city to its bus stations were open. "The correct option is B. No major roads connecting the city to its bus stations were open. For the statement :All integers are not numbers. The correct option is D. All integers are not numbers. For the part C :The correct option is A. A∪U={1,2,3,4,5,6,7}.
Part A:The given statement is "No major roads connecting the city to its bus stations were open."A. At least one major road connecting the city to its bus stations was openB. No major roads connecting the city to its bus stations were open.C. The bus stations were forced to closeD. At least one major road connecting the city to its bus stations was not open.The correct option is B. No major roads connecting the city to its bus stations were open.
Part B:All integers are not numbers. We need to express the quantified statement in an equivalent way and then write the negation of the quantified statement. The equivalent statement of the quantified statement is "Not all integers are numbers". A. No integer is a number. B. All integers are not numbers. C. Not all integers are numbers. D. At least one integer is a number. The correct option is C. Not all integers are numbers. The negation of the quantified statement is "All integers are numbers". A. Some integers are not numbers. B. Some integers are numbers. C. No integers are numbers. D. All integers are not numbers. The correct option is D. All integers are not numbers.
Part C:U={1,2,3,4,5,6,7,} and A={5,6,7,6}.We need to find union A∪U.A∪U = {1,2,3,4,5,6,7}The correct option is A. A∪U={1,2,3,4,5,6,7}.
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Q3: Let X ~ N (µ₁, 0₁²), and Y ~ N (µ₂, 0₂²). Furthermore, X and Y are independent to each other. Please derive the distribution of (1) X+Y (2) X-Y (3) 4X-2Y+6
The distribution of 4X - 2Y + 6 is given by 4X - 2Y + 6 ~ N(4µ₁ - 2µ₂ + 6, 16(0₁²) + 4(0₂²)).
To derive the distributions of (1) X+Y, (2) X-Y, and (3) 4X-2Y+6, we can use the properties of normal distributions and the fact that X and Y are independent.
(1) X + Y:
Since X and Y are independent normal random variables, their sum follows a normal distribution. The mean of the sum is the sum of the means, and the variance of the sum is the sum of the variances.
Mean of X + Y: µ₁ + µ₂
Variance of X + Y: 0₁² + 0₂²
Therefore, the distribution of X + Y is given by X + Y ~ N(µ₁ + µ₂, 0₁² + 0₂²).
(2) X - Y:
Similar to (1), the difference of two independent normal random variables also follows a normal distribution. The mean of the difference is the difference of the means, and the variance of the difference is the sum of the variances.
Mean of X - Y: µ₁ - µ₂
Variance of X - Y: 0₁² + 0₂²
Thus, the distribution of X - Y is given by X - Y ~ N(µ₁ - µ₂, 0₁² + 0₂²).
(3) 4X - 2Y + 6:
To find the distribution of this expression, we can apply the properties of linear combinations of normal random variables. The mean and variance of a linear combination can be calculated by multiplying the mean and variance of each random variable by their respective coefficients and summing them up.
Mean of 4X - 2Y + 6: 4µ₁ - 2µ₂ + 6
Variance of 4X - 2Y + 6: (4²)(0₁²) + (-2²)(0₂²) = 16(0₁²) + 4(0₂²) = 16(0₁²) + 4(0₂²)
Hence, the distribution of 4X - 2Y + 6 is given by 4X - 2Y + 6 ~ N(4µ₁ - 2µ₂ + 6, 16(0₁²) + 4(0₂²)).
Please note that in each case, the resulting distribution is still a normal distribution with a new mean and variance based on the properties of linear combinations and the assumptions of independence.
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(Simple Regression - Calculation - 30) A research team collected data on 201 students in a statistics course. Their dependent variable (response Y ) was the student's score on the final examination, which ranged from 200 to 600 points. The observed average final examination score was 452 , with an observed standard deviation of 35.5 (the divisor was n−1=200 ). Their independent variable (predictor x ) was the score on the first examination in the course, which also ranged from 200 to 600 . The average was 480 , with an observed standard deviation of 50.3. The correlation coefficient between the first examination score and the final examination score was 0.75. If we fit a simple linear model Y i
∼N(β 0
+β 1
x i
,σ 2
),i=1,…,n=201. (You can use the results in HW2-Q4 directly.) (a) (10points) Report the ANOVA table for the model. (b) (10points) Test the null hypothesis H 0
:β 1
=0 v.s. H 1
:β 1
=0, with 0.05 level of significance. (c) Find the confidence interval for the expected final examination score of students who scored 600 on the first examination. (d) Find the prediction interval for the final examination score of a student who scored 600 on the first examination.
The interval is[tex]$632.74 \pm 1.972 \times 10.275 \times \sqrt{1 + \frac{1}{201} + \frac{(600 - 480)^2}{\sum(x_i - \overline{x})^2}} = [541.38, 724.11]$.[/tex]
a) The table is as follows:
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
Source & Degrees of Freedom & Sum of Squares & Mean Square & F Value & Pr > F \\
\hline
Model & 1 & 26697.66 & 26697.66 & 639.27 & $<.0001$ \\
Error & 199 & 8315.62 & 41.74 & & \\
\hline
\end{tabular}
\end{center}
b) The null hypothesis is [tex]$H_0 : \beta_1 = 0$ vs $H_1 : \beta_1 \neq 0$. The t-statistic is given by:\[t = \frac{0.75 - 0}{\left(\frac{35.5}{\sqrt{201}}\right) / \left(\frac{50.3}{\sqrt{201}}\sqrt{1 - 0.75^2}\right)} = 13.27.\][/tex]
Since the degrees of freedom are $n - 2 = 201 - 2 = 199$, the two-tailed p-value is less than 0.0001. Hence, we reject the null hypothesis. Therefore, there is significant evidence that the slope of the regression line is nonzero.
c) The 95% confidence interval is given by:
[tex]\[y_0 \pm t_{0.025,199}\,s[\varepsilon]\sqrt{\frac{1}{n} + \frac{(x_0 - \overline{x})^2}{\sum(x_i - \overline{x})^2}},\]\\[/tex]
where [tex]$y_0 = \beta_0 + \beta_1x_0 = 299.04 + 0.5669 \times 600 = 632.74$, $t_{0.025,199} = 1.972$, $s[\varepsilon] = \sqrt{\frac{8315.62}{199}} = 10.275$, $x_0 = 600$, and $\overline{x} = 480$[/tex]. Therefore, the interval is [tex]$632.74 \pm 1.972 \times 10.275 \times \sqrt{\frac{1}{201} + \frac{(600 - 480)^2}{\sum(x_i - \overline{x})^2}} = [609.29, 656.19]$.[/tex]
d) The 95% prediction interval is given by:
[tex]\[y_0 \pm t_{0.025,199}\,s[\varepsilon]\sqrt{1 + \frac{1}{n} + \frac{(x_0 - \overline{x})^2}{\sum(x_i - \overline{x})^2}},\]\\where $t_{0.025,199} = 1.972$,[/tex] and all the other variables have been defined in part c. Therefore, the interval is [tex]$632.74 \pm 1.972 \times 10.275 \times \sqrt{1 + \frac{1}{201} + \frac{(600 - 480)^2}{\sum(x_i - \overline{x})^2}} = [541.38, 724.11]$.[/tex]
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Given a language L, the complement is defined as Lˉ={w∣w∈Σ∗ and w∈/L}. Given a language L, a DFA M that accepts L is minimal if there does not exist a DFA M′ such that M′ accepts L, and M′ has fewer states than M. (a) Prove that the class of regular languages is closed under complement. (b) Given a DFA M that accepts L, define Mˉ to be the DFA that accepts Lˉ using your construction from part (a). Prove that if M is minimal, then Mˉ is minimal.
If M is a minimal DFA accepting L, then the DFA Mˉ accepting the complement of L is also minimal.
(a) To prove that the class of regular languages is closed under complement, we need to show that for any regular language L, its complement Lˉ is also a regular language.
Let's assume that L is a regular language. This means that there exists a DFA (Deterministic Finite Automaton) M that accepts L. We need to construct a DFA M' that accepts the complement of L, Lˉ.
To construct M', we can simply swap the accepting and non-accepting states of M. In other words, for every state q in M, if q is an accepting state in M, then it will be a non-accepting state in M', and vice versa. The transition function and start state remain the same.
The intuition behind this construction is that M accepts strings that are in L, and M' will accept strings that are not in L. By swapping the accepting and non-accepting states, M' will accept the complement of L.
Since we can construct a DFA M' that accepts Lˉ from the DFA M that accepts L, we have shown that Lˉ is a regular language. Therefore, the class of regular languages is closed under complement.
(b) Now, let's assume that M is a minimal DFA that accepts the language L. We need to prove that Mˉ, the DFA accepting the complement of L, is also minimal.
To prove this, we can use a contradiction argument. Let's assume that Mˉ is not minimal, i.e., there exists a DFA M'' that accepts Lˉ and has fewer states than M. Our goal is to show that this assumption leads to a contradiction.
Since M is minimal, it means that there is no DFA M' that accepts L and has fewer states than M. However, we have assumed the existence of M'', which accepts Lˉ and has fewer states than M.
Now, consider the DFA M''', obtained by swapping the accepting and non-accepting states of M''. In other words, for every state q in M'', if q is an accepting state in M'', then it will be a non-accepting state in M''', and vice versa. The transition function and start state remain the same.
We can observe that M''' accepts L because it accepts the complement of Lˉ, which is L. Moreover, M''' has fewer states than M, which contradicts the assumption that M is minimal.
Therefore, our initial assumption that Mˉ is not minimal leads to a contradiction. Hence, if M is minimal, then Mˉ is also minimal.
In conclusion, we have proven that if M is a minimal DFA accepting L, then the DFA Mˉ accepting the complement of L is also minimal.
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Suppose your aumt bought a new car for $10,500 in 2012 , and that the value of the car depreciates by $600 each year. Find the function V(t) that gives the value of the car in dollars; where t is the number of years since 2012 . V(t)= Accorting to the model, the vatue of your aunt's car in 2017 was ________.
The value of your aunt's car in 2017, according to the given model, was $7,500.
To find the function V(t) that gives the value of the car in dollars, we start with the initial value of the car in 2012, which is $10,500. Since the car depreciates by $600 each year, the value decreases by $600 for every year elapsed.
We can express the function V(t) as follows:
V(t) = 10,500 - 600t
where t represents the number of years since 2012.
To find the value of your aunt's car in 2017, we substitute t = 5 (since 2017 is 5 years after 2012) into the function:
V(5) = 10,500 - 600 * 5
= 10,500 - 3,000
= $7,500
Therefore, the value of your aunt's car in 2017, according to the given model, was $7,500.
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f f ( 2 ) = 14 and f ' ( x ) ≥ 1 for 2 ≤ x ≤ 7 , how small can f ( 7 ) possibly be?
Answer:
f(7) ≥ 19
Step-by-step explanation:To find the smallest possible value of f(7), we can use the Mean Value Theorem for Derivatives. According to this theorem, if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
In this case, we know that f(2) = 14 and f'(x) ≥ 1 for 2 ≤ x ≤ 7. Therefore, we can apply the Mean Value Theorem to the interval [2, 7] to get:
f'(c) = (f(7) - f(2))/(7 - 2)
Since f'(x) ≥ 1 for 2 ≤ x ≤ 7, we have:
1 ≤ f'(c) = (f(7) - 14)/5
Multiplying both sides by 5 and adding 14, we get:
f(7) ≥ 19
The response to a question has three altematives: A, B, and C. A sample of 120 responses provides 62 A,24 B, and 34C responses. Show the frequency and relative frequency distributions (use nearest whole number for the frequency column and 2 decimal for the relative frequency column).
Alternatives Response Frequency Relative Frequency of A62/120 = 0.52 Relative Frequency of B24/120 = 0.20 Relative Frequency of C34/120 = 0.28 Total 120/120 = 1
Given that there are 3 alternatives to the answer of a question, A, B, and C. In a sample of 120 responses, there are 62 A, 24 B, and 34 C responses. We are required to create the frequency and relative frequency distributions for the given data. Frequency distribution Frequency distribution is defined as the distribution of a data set in a tabular form, using classes and frequencies. We can create a frequency distribution using the given data in the following manner: Alternatives Response Frequency Frequency of A62 Frequency of B24 Frequency of C34 Total 120
Thus, the frequency distribution table is obtained. Relationship between the frequency and the relative frequency: Frequency is defined as the number of times that a particular value occurs. It is represented as a whole number or an integer. Relative frequency is the ratio of the frequency of a particular value to the total number of values in the data set. It is represented as a decimal or a percentage. It is calculated using the following formula: Relative frequency of a particular value = Frequency of the particular value / Total number of values in the data set Let us calculate the relative frequency of the given data:
Alternatives Response Frequency Frequency of A62 Frequency of B24 Frequency of C34 Total 120 Now, we can calculate the relative frequency as follows:
Alternatives Response Frequency Relative Frequency of A62/120 = 0.52Relative Frequency of B24/120 = 0.20Relative Frequency of C34/120 = 0.28 Total 120/120 = 1 The relative frequency distribution table is obtained.
We have calculated the frequency and relative frequency distributions for the given data. The frequency distribution is obtained using the classes and frequencies, and the relative frequency distribution is obtained using the ratio of the frequency of a particular value to the total number of values in the data set.
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the volume v of a melting snowball is decreasing at at rate of 4 cm3 per second. let the variable t represent the time, in seconds, since we started our investigation. find the rate at which the radius of the snowball is decreasing with respect to time at the instant when the radius of the snow ball is 3 . round your answer to three decimal place accuracy.
The rate at which the radius of the snowball is decreasing with respect to time is approximately 0.035 cm/s when the radius of the snowball is 3 cm.
How to calculate the rate
Volume V of the snowball to its radius r is given by
[tex]V = (4/3) \pi r^3[/tex]
Take the derivative of both sides with respect to time t, we get:
[tex]dV/dt = 4\pi r^2 (dr/dt)[/tex]
where dr/dt is the rate at which the radius is changing with respect to time.
[tex]dV/dt = -4 cm^3/s[/tex] (negative because the volume is decreasing),
To find dr/dt when the radius is 3 cm.
substitute these values and solve for dr/dt:
[tex]-4 cm^3/s = 4\pi (3 cm)^2 (dr/dt)[/tex]
[tex]dr/dt = (-4 cm^3/s) / (36\pi cm^2) = -0.035 cm/s[/tex]
Thus, the rate at which the radius of the snowball is decreasing with respect to time is approximately 0.035 cm/s when the radius of the snowball is 3 cm.
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Let f(t) denote the number of people eating in a restaurant & minutes after 5 PM. Answer the following questions:
a) Which of the following statements best describes the significance of the expression f(4) = 177
A. Every 4 minutes, 17 more people are eating
B. There are 17 people eating at 9:00 PM
C. There are 4 people eating at 5:17 PM
D. There are 17 people eating at 5:04 PM
E. None of the above
b) Which of the following statements best describes the significance of the expression f(a) = 26?
A, a minutes after 5 PM there are 26 people eating
B. Every 26 minutes, the number of people eating has increased by a people
C. At 5:26 PM there are a people eating
D. a hours after 5 PM there are 26 people eating
E. None of the above
c) Which of the following statements best describes the significance of the expression f(26) = b?
A. Every 26 minutes, the number of people eating has increased by b people
B. 6 hours after 5 PM there are 26 people eating
c. At 5:26 PM there are & people eating
D. 6 minutes after 5 PM there are 26 people eating
E. None of the above
d) Which of the following statements best describes the significance of the expression n
A. f hours after 5 PM there are 7 people eating,f(t)?
B. Every f minutes, r more people have begun eating
C. n hours after 5 PM there are t people eating
D. 7 minutes after 5 PM there are t people eating
E. None of the above
For (a) none of the given options accurately describe the significance of the expression and for (b) option A is the answer.
The statement "f(4) = 177" means that there are 177 people eating in the restaurant 4 minutes after 5 PM. Therefore, none of the given options accurately describe the significance of the expression.
The statement "f(a) = 26" means that a minutes after 5 PM, there are 26 people eating in the restaurant. Therefore, option A, "a minutes after 5 PM there are 26 people eating," best describes the significance of the expression.
The given expressions represent the number of people eating in the restaurant at different points in time. By substituting specific values into the function f(t), we can determine the number of people eating at a particular time. It is important to note that without additional context or information about the function f(t) or the behavior of the restaurant's patrons, we cannot make definitive conclusions about the exact number of people eating at specific times. The given expressions only provide information about the number of people at specific time intervals or with specific variables.
In summary, the expressions f(t) represent the number of people eating in the restaurant at different times. The significance of each expression depends on the specific values provided or the relationships between variables, and without more information, it is challenging to draw precise conclusions about the exact number of people at specific times.
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Real Analysis
Prove that for all natural numbers \( n, 2^{n-1} \leq n ! \). (Hint: Use induction)
To prove the inequality [tex]\(2^{n-1} \leq n!\)[/tex] for all natural numbers \(n\), we will use mathematical induction.
Base Case:
For [tex]\(n = 1\)[/tex], we have[tex]\(2^{1-1} = 1\)[/tex] So, the base case holds true.
Inductive Hypothesis:
Assume that for some [tex]\(k \geq 1\)[/tex], the inequality [tex]\(2^{k-1} \leq k!\)[/tex] holds true.
Inductive Step:
We need to prove that the inequality holds true for [tex]\(n = k+1\)[/tex]. That is, we need to show that [tex]\(2^{(k+1)-1} \leq (k+1)!\).[/tex]
Starting with the left-hand side of the inequality:
[tex]\(2^{(k+1)-1} = 2^k\)[/tex]
On the right-hand side of the inequality:
[tex]\((k+1)! = (k+1) \cdot k!\)[/tex]
By the inductive hypothesis, we know that[tex]\(2^{k-1} \leq k!\).[/tex]
Multiplying both sides of the inductive hypothesis by 2, we have [tex]\(2^k \leq 2 \cdot k!\).[/tex]
Since[tex]\(2 \cdot k! \leq (k+1) \cdot k!\)[/tex], we can conclude that [tex]\(2^k \leq (k+1) \cdot k!\)[/tex].
Therefore, we have shown that if the inequality holds true for \(n = k\), then it also holds true for [tex]\(n = k+1\).[/tex]
By the principle of mathematical induction, the inequality[tex]\(2^{n-1} \leq n!\)[/tex]holds for all natural numbers [tex]\(n\).[/tex]
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what is the slope of the line that contains thenpoints (6,0)(0,3) and (12,-3)?
The slope of the line passing through the points (6,0), (0,3), and (12,-3) is -0.5.
The slope of a line passing through two points, we can use the formula: slope (m) = (change in y) / (change in x). We will use the points (6,0) and (0,3) to calculate the slope.
1. Calculate the change in y:
Δy = y₂ - y₁ = 0 - 3 = -3
2. Calculate the change in x:
Δx = x₂ - x₁ = 6 - 0 = 6
3. Substitute the values into the slope formula:
m = Δy / Δx = -3 / 6 = -0.5
Therefore, the slope of the line passing through the points (6,0) and (0,3) is -0.5. It is worth noting that the third point (12,-3) was not used in the calculation of the slope, as the slope remains the same regardless of the additional point.
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tra Credit] The function \( f:[0, \pi / 2] \rightarrow[-1,1] ; f(x)=\cos (x) \) is: decreasing injective surjective none of these properties invertible increasing
The function \( f(x) = \cos(x) \) on the interval \([0, \pi/2]\) is decreasing.
The cosine function is defined as the ratio of the adjacent side to the hypotenuse in a right triangle. As the angle increases from 0 to \(\pi/2\), the adjacent side decreases while the hypotenuse remains constant, resulting in a decreasing function. This can also be observed from the graph of the cosine function, where it starts at its maximum value of 1 at \(x = 0\) and decreases continuously until it reaches its minimum value of -1 at \(x = \pi/2\).
In terms of injectivity, the cosine function is not injective on the interval \([0, \pi/2]\). Injectivity means that each element in the domain is mapped to a unique element in the range. However, since the cosine function is periodic with a period of \(2\pi\), multiple values of \(x\) can produce the same value of \(\cos(x)\). For example, both \(x = 0\) and \(x = 2\pi\) result in \(\cos(x) = 1\).
Regarding surjectivity, the cosine function is surjective on the interval \([-1, 1]\), which means that for any given value \(y\) in the range \([-1, 1]\), there exists at least one value \(x\) in the domain \([0, \pi/2]\) such that \(f(x) = y\). This is because the cosine function oscillates between -1 and 1 infinitely, covering the entire range \([-1, 1]\) within the interval \([0, \pi/2]\).
Based on the above explanations, the correct answer is that the function \(f(x) = \cos(x)\) is decreasing.
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Use the following information to fill in the the statements below. The graph on the right shows a sample of 325 observations from a population with unknown μ. Using this information, which of the following best describes the true sampling distribution of the sample mean. Histogram of the Sample Data 1.95 2.00 sample data 50 40 30 Frequency 20 10 T 1.85 1.90 2.05 According to the Central Limit Theorem, the shape of the distribution of sample means will b✓ [Select] because the [Select] exponential uniform normal bimodal According to the Central Limit morem, the standard deviation of the distribution of According to the Central Limit Theorem, the shape of the distribution of sample means will be [Select] because the [Select] standard deviation is greater than 1 standard deviation is considered large enough. population mean is not known sample size is considered large enough According to the Central Limit Theorem, the standard deviation of the distribution of [Select] According to the Central Limit Theorem, the standard deviation of the distribution of the sample mean✓ [Select] always smaller than the standard deviation of the population is always larger than the standard deviation of the population equal to the population standard deviation.
According to the information provided, the correct answers are as follows:
1. The shape of the distribution of sample means will be normal because the population mean is not known and the sample size is considered large enough.
2. The standard deviation of the distribution of the sample mean is always smaller than the standard deviation of the population.
1. According to the Central Limit Theorem, when the sample size is large enough, regardless of the shape of the population distribution, the distribution of sample means tends to follow a normal distribution.
2. The standard deviation of the distribution of the sample mean, also known as the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size. Since the sample mean is an average of observations, the variability of the sample mean is reduced compared to the variability of individual observations in the population.
The Central Limit Theorem states that when the sample size is sufficiently large, the distribution of sample means will be approximately normal, regardless of the shape of the population distribution. The standard deviation of the sample mean will be smaller than the standard deviation of the population.
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Use the Product Rule to evaluate and simplify d/dx((x-3)(4x+2)).
Answer:
8x - 10
Step-by-step explanation:
Let [tex]f(x)=x-3[/tex] and [tex]g(x)=4x+2[/tex], hence, [tex]f'(x)=1[/tex] and [tex]g'(x)=4[/tex]:
[tex]\displaystyle \frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)=1(4x+2)+(x-3)\cdot4=4x+2+4(x-3)=4x+2+4x-12=8x-10[/tex]
A surgeon tells you that for every 150 surgeries that she perfos, 6 patients need to come back for the second surgery. If you are the next patient, what is the probability that you would need to have the second surgery? Round your answer to the nearest hundredth.
The probability that the patient would need to have the second surgery is 0.04 or 4% rounded to the nearest hundredth.
Given that for every 150 surgeries a surgeon performs, 6 patients need to come back for the second surgery. According to the given data, the probability that a patient would need to have the second surgery can be determined as follows:
Probability of not needing the second surgery:
P(not needing the second surgery) = 1 - P(needing the second surgery)
P(not needing the second surgery) = 1 - 6/150P(not needing the second surgery)
= 1 - 0.04P(not needing the second surgery)
= 0.96
Probability of needing the second surgery:
P(needing the second surgery) = 6/150P(needing the second surgery)
= 0.04
Therefore, the probability that the patient would need to have the second surgery is 0.04 or 4% rounded to the nearest hundredth.
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Leo buy a robot during the ale. If the original price wa $180, how much doe Leo pay?
Leo pays $135 for the robot after applying a 25% discount.
To calculate how much Leo pays for the robot after applying a 25% discount, we can use the following formula:
Amount paid = Original price - (Discount percentage × Original price)
Given that the original price of the robot is $180 and the discount percentage is 25% (0.25), we can substitute these values into the formula:
Amount paid = $180 - (0.25 × $180)
Calculating the expression:
Amount paid = $ 180 - ($45)
Amount paid = $ 135
Therefore, Leo pays $135 for the robot after applying a 25% discount.
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Complete question is :
Leo buy a robot during the sale at 25 % discount. If the original price was $180, how much doe Leo pay?
a group of 95 students were surveyed about the courses they were taking at their college with the following results: 57 students said they were taking math. 57 students said they were taking english. 62 students said they were taking history. 32 students said they were taking math and english. 39 students said they were taking math and history. 36 students said they were taking english and history. 19 students said they were taking all three courses. how many students took none of the courses?
Out of the 95 students surveyed, 7 students took none of the courses. To find the number of students who took none of the courses, we need to subtract the number of students who took at least one course from the total number of students surveyed.
First, let's find the number of students who took at least one course. We can do this by adding the number of students who took each course individually, and then subtracting the students who took two courses and the students who took all three courses.
The number of students who took math is 57, the number who took English is 57, and the number who took history is 62. To find the total number of students who took at least one course, we add these numbers: 57 + 57 + 62 = 176.
Now, we need to subtract the number of students who took two courses. We know that 32 students took math and English, 39 students took math and history, and 36 students took English and history. To find the total number of students who took two courses, we add these numbers: 32 + 39 + 36 = 107.
Next, we need to subtract the number of students who took all three courses. We know that 19 students took all three courses.
To find the number of students who took none of the courses, we subtract the students who took at least one course (176) from the students who took two courses (107) and the students who took all three courses (19):
95 - 176 + 107 - 19 = 7.
Therefore, the number of students who took none of the courses is 7.
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Find the standard form for the equation of a circle (x-h)^(2)+(y-k)^(2)=r^(2) with a diameter that has endpoints (-6,1) and (10,8)
The standard form of the equation of a circle with a diameter that has endpoints (-6,1) and (10,8) is
[tex](x - 2)^2 + (y - 4.5)^2 = 64[/tex].
To find the standard form of the equation of a circle, we need to determine the center coordinates (h, k) and the radius (r).
First, we find the midpoint of the line segment connecting the endpoints of the diameter. The midpoint formula is given by:
[tex]\[ \left( \frac{{x_1 + x_2}}{2}, \frac{{y_1 + y_2}}{2} \right) \][/tex]
Using the coordinates of the endpoints (-6,1) and (10,8), we calculate the midpoint as:
[tex]\[ \left( \frac{{-6 + 10}}{2}, \frac{{1 + 8}}{2} \right) = (2, 4.5) \][/tex]
The coordinates of the midpoint (2, 4.5) represent the center (h, k) of the circle.
Next, we calculate the radius (r) of the circle. The radius is half the length of the diameter, which can be found using the distance formula:
[tex]\[ \sqrt{{(x_2 - x_1)^2 + (y_2 - y_1)^2}} \][/tex]
Using the coordinates of the endpoints (-6,1) and (10,8), we calculate the distance as:
[tex]\[ \sqrt{{(10 - (-6))^2 + (8 - 1)^2}} = \sqrt{{256 + 49}} \\\\= \sqrt{{305}} \][/tex]
Therefore, the radius (r) is [tex]\(\sqrt{{305}}\)[/tex].
Finally, we substitute the center coordinates (2, 4.5) and the radius [tex]\(\sqrt{{305}}\)[/tex]into the standard form equation of a circle:
[tex]\[ (x - 2)^2 + (y - 4.5)^2 = (\sqrt{{305}})^2 \][/tex]
Simplifying and squaring the radius, we get:
[tex]\[ (x - 2)^2 + (y - 4.5)^2 = 64 \][/tex]
Hence, the standard form of the equation of the circle is [tex](x - 2)^2 + (y - 4.5)^2 = 64.[/tex]
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