can anyone give there snap?

The curve r = 5 + 4 cos(theta) the area enclosed by the curve is 32.5π square units.

The curve you've provided is given by the polar equation r = 5 + 4 cos(theta). This curve represents a limaçon, a specific type of polar curve.

To find the area enclosed by the curve, you can use the polar area formula: Area = (1/2) ∫[r^2 d(theta)], where the integral is evaluated over the range of theta for one full rotation.

In this case, r = 5 + 4 cos(theta), and theta ranges from 0 to 2π: Area = (1/2) ∫[(5 + 4 cos(theta))^2 d(theta)] from 0 to 2π. Evaluating this integral, we get: Area = (1/2) * (65π) = 32.5π square units.

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Therefore, the correct is: 2 sin(3πft) where f' is the cut-off frequency of the low-pass filter.

The sampling interval is T = 0.5 seconds, since there are two samples per second.

The Nyquist frequency is equal to half of the sampling frequency, or f_Nyquist = 1 / (2T) = 1 Hz. Since the frequency of the original signal is 2rf = 2(2πf) = 4πf, which is greater than the Nyquist frequency, the signal will be aliased.

The aliased signal can be obtained by subtracting the Nyquist frequency from the original frequency, which gives 4πf - 2πf_Nyquist = 4πf - π = 3πf. Therefore, the aliased signal has frequency 3πf.

The sampled signal can be expressed as:

x(nT) = 2 sin(2πfnT)

where n is an integer representing the sample number. The recovered signal can be obtained by applying a low-pass filter to this signal to remove the high-frequency component due to aliasing. The cut-off frequency of the filter should be less than or equal to the Nyquist frequency to ensure that no aliasing occurs.

The recovered signal can be expressed as:

x_recovered(t) = 2 sin(2πf' t)

where f' is the cut-off frequency of the low-pass filter. Since the original signal has frequency 4πf, the recovered signal should have frequency 3πf to avoid aliasing.

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Complete question:

The signal is in blue, is sampled regularly, with the red dots indicating the sample values. what is the signal that will be recovered from the sample values?

The series -2/3 - 4/4 + 6/5 + 8/6 - 10/7 + ... is divergent and the series is in the form ∑ [tex]bn = b1 + b2 + b3 + ...,[/tex]  where bn is the nth term of the series.

To distinguish bn, we need to compose the given series within the form:

[tex]bn = b1 + b2 + b3 + ...[/tex]

where bn is the nth term of the series.

Looking at the given arrangement, we see that the numerators of the terms are substituting indeed and odd integrability, beginning with 2 and expanding by 2 for each indeed term and diminishing by 1 for each odd term.

The denominators are basically the integers 3, 4, 4, 5, 6, 6, 7, ...

So, ready to type in the nth term of the arrangement as:

[tex]bn = (-1)^{2} (n+1) * (2n - 1) / (n + 2)[/tex]

Presently, we are able to test for meeting or uniqueness utilizing the substituting arrangement test.

The rotating arrangement test states that on the off chance that an arrangement fulfills the taking-after conditions:

The terms substitute in sign.

The absolute esteem of each term is diminishing.

The constraint of the absolute esteem of the terms as n approaches boundlessness is zero.

At that point, the series converges.

In our case, the terms interchange in sign, and we are able to appear that the absolute value of each term is diminishing as takes after:

[tex]|bn+1| = (2n + 1) / (n + 3) < (2n - 1) / (n + 2) = |bn|[/tex]

So, the moment condition is fulfilled.

To appear that the third condition is fulfilled, we will take the restrain of the supreme value of bn as n approaches infinity:

lim (n→∞) |bn| = lim (n→∞) (2n - 1) / (n + 2) = 2

Since the constraint isn't zero, the rotating arrangement test does not apply, and we cannot conclude whether the arrangement merges or veers based on that test alone.

Instep, we will utilize the constrain comparison test. Let's compare our arrangement to the arrangement ∑(1/n) by taking the restrain of the proportion of the nth terms:

lim (n→∞) |bn| / (1/n) = lim (n→∞) n(2n - 1) / (n + 2)

Isolating the numerator and denominator by[tex]n^2,[/tex]we get:

lim (n→∞) |bn| / (1/n) = lim (n→∞) (2 - 1/n) / (1 + 2/n)

Since both the numerator and denominator approach constants as n approach infinity, we will take the limit as n approaches infinity directly and get:

lim (n→∞) |bn| / (1/n) = 2

This implies that our arrangement and the arrangement ∑(1/n) carry on additionally in the limit, and since the consonant arrangement ∑(1/n) diverges, able to conclude that our series also diverges by the limit comparison test.

Hence, the series -2/3 - 4/4 + 6/5 + 8/6 - 10/7 + ... diverges.

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The value after one year will be $535.

To explain in the simplest form, interest is calculated as a percent of the principal. For example, assume that you have borrowed $100 from your friend and you have promised to repay it with 5% interest, then the amount of interest you would pay along with the actual amount would just be 5% of 100 which is $100(5/100) = $5.

An annual percentage of the amount of a loan is known as interest. For example, when you deposit your money in a high-yield savings account, the bank will pay interest. Now, according to the question

Given the amount = $500

interest rate is given as 7% on 3-year certificates of deposit.

Therefore, the value after one year will be

= 500 x 7% + 500

=500 x 0.07 + 500

= 35 + 500

= $535

Hence, the value will be $535.

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The requreid local baseball team sold 99 adult tickets, 66 child tickets, and 22 senior tickets for the game.

Let A, C, and S represent the number of adult, child, and senior tickets sold, respectively.

A + C + S = 187 (the total number of tickets sold)

A:C = 3:2 (the ratio of adult to child tickets)

A:S = 9:2 (the ratio of adult to senior tickets)

We can use the ratios to write:

A = 3x (where x is a common factor)

C = 2x

S = (2/9)A = (2/9)(3x) = (2/3)x

Now we can substitute these expressions into the first equation:

A + C + S = 3x + 2x + (2/3)x = (9/3)x + (6/3)x + (2/3)x = 17x/3 = 187

x = 187(3/17) ≈ 33

Therefore, we can find the number of adults, children, and senior tickets sold by multiplying x by the appropriate ratio factors:

A = 3x ≈ 99

C = 2x ≈ 66

S = (2/3)x ≈ 22

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Using the equation

92001 - 1600 x n(t) x 4 + t

A)

There will be about 9041 infected people by the end of the 6th month.

B)

There are no 6400 infected people according to this model.

C)

There will not be more than 8800 infected people.

We have,

(a)

To find the number of people infected by the end of the 6th month, we need to substitute t = 6 into the formula.

= 92001 - 1600 n(6) 4+6

= 92001 - 1600 n(6) 10

= 92001 - 160000

= 9041

(b)

To find the time when there are 6400 infected people, we need to solve the equation:

92001 - 1600 n(t) 4+t = 6400

1600 n(t) 4+t = 85601

n(t) = 85601 / (1600 (4+t))

We need to solve for t when n(t) = 6400:

6400 = 85601 / (1600 (4+t))

4+t = 85601 / (6400 × 1600) ≈ 0.84

t ≈ 0.84 - 4 ≈ -3.16

Since time cannot be negative, we can conclude that there are no 6400 infected people according to this model.

(c)

We need to find if n(t) > 8800 for all t > 0. We can check by evaluating n(t) at t = 0 and at a large value of t.

n(0) = 92001 - 1600 × 0 × 4+0 = 92001

n(100) = 92001 - 1600 × 100 × 4+100 = - 639999

Since n(100) is negative, we can conclude that according to this model, there will not be more than 8800 infected people.

Thus,

A)

There will be about 9041 infected people by the end of the 6th month.

B)

There are no 6400 infected people according to this model.

C)

There will not be more than 8800 infected people.

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The number of sharks found between 76 m depth and 100 m depth is 2.04 shark's population.

The number of sharks found at different depths is calculated as follows;

From the histogram, the population of the fish at depth between 76 m and 100 m can be read off by tracing the depth values towards the frequency axis;

at depth 76 m, the frequency = 1.5

at depth 100 m, the frequency = 0.54

Total number of fish between the depths = 1.5 + 0.54 = 2.04

Thus, this value represents the density of the sharks between 76 m and 100 m.

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The normal curve may be used to estimate the sampling distribution for alternatives (B) and (C) because they have sample sizes and population characteristics.

When the sample size is sufficient and the success probability (p) is not too near 0 or 1, the normal approximation to the binomial distribution can be employed.

The normal approximation is suitable, according to a widely accepted rule of thumb, when both np and n(1-p) are higher than or equal to 10.

Let's examine the available options:

If (A) n=15 and p=0.6, np=15 * 0.6 = 9, and n(1-p)=15 * 0.4 = 6, both are less than 10, preventing the adoption of the normal approximation.

When n=30 and p=0.3, the normal approximation may be employed since np=30 * 0.3 = 9 and n(1-p)=30 * 0.7 = 21 both are higher than or equal to 10.

When n=30 and p=0.6, the normal approximation may be employed since np=30 * 0.6 = 18 and n(1-p)=30 * 0.4 = 12 are both higher than or equal to 10.

(D) If n=15 and p=0.3, np=15*0.3 =4.5 and n(1-p)=15*0.7 =10.5, respectively; np is less than 10, therefore the typical approximation cannot be applied.

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The vertical asymptote is x = -4 and the horizontal asymptote of the function is y = 2.

To find the vertical asymptote of the function f(x) = 2x ÷ (x+4), we need to look for any value of x that makes the denominator equal to zero. In this case, we have: x + 4 = 0

x = -4

Therefore, the vertical asymptote is x = -4.

f(x) = (2x ÷ x) ÷ (x ÷ x + 4 ÷ x)

f(x) = 2 ÷ (1 + 4/x)

As x becomes very large, the term 4/x becomes very small and can be neglected.

Therefore, as x → ∞, f(x) → 2/1 = 2.

Similarly, as x becomes very small (i.e., negative), the term 4/x becomes very large and can be neglected. Therefore, as

x → -∞, f(x) → 2/1 = 2.

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The complete question is:

Determine the equations of the vertical and horizontal asymptotes, if any, for f(x) = 2x / x + 4.

Answer:

If the perimeter of a square picture frame is 12 inches, then each side of the square frame must be 3 inches long, since 4 x 3 = 12.

To find the area of the picture frame, we need to subtract the area of the picture from the area of the frame. Since the frame is a square with 3-inch sides, its area is 3 x 3 = 9 square inches.

However, we don't know the size of the picture, so we can't calculate its area directly. Instead, we can use the fact that the picture and the frame together form a larger square with sides that are 12 inches long (since the perimeter of the whole thing is 12 inches).

The area of this larger square is 12 x 12 = 144 square inches.

Since the area of the frame is 9 square inches, the area of the picture must be 144 - 9 = 135 square inches.

Therefore, the area of the picture frame is 9 square inches, and the area of the picture is 135 square inches.

Answer:

Anna will get 350 ZED

Step-by-step explanation:

since jenny is getting 3/8ths of the money, we can find how much money Jenny is getting and subtract that amount from the original total. to find this, take the original amount divided by the denominator then multiplied by the numerator.

for example: 560 / 8 = 70 × 3 = 210

560 - 210 = 350

350 is how much anna will get.

(a)  (AB)1,2 = (1)(-1) + (2)(3) + (-2)(1) = -1 + 6 - 2 = 3. (b)  (AB)2,3 and (AB)3,2 do not exist. (c) To determine if BA exists, we need to check if the number of columns in matrix B is equal to the number of rows in matrix A. B has 2 columns and A has 4 rows, so BA does not exist. (d) Since we don't have matrix C, we cannot find CA.

(a) To find (AB)1,2 without computing the whole matrix, we only need to compute the dot product of the first row of matrix A and the second column of matrix B.
A = | 1  2 |
   |-2  3 |
   | 2  4 |
   |10  4 |
B = | 3 -1 |
   | 1  5 |
   | 3  1 |
   | 2  2 |
(AB)1,2 = (1 * -1) + (2 * 5) = -1 + 10 = 9
(b) (AB)2,3 and (AB)3,2 do not exist because matrix A has 2 columns and matrix B has 3 rows. For these elements to exist, matrix A should have 3 columns and matrix B should have 3 rows.
(c) BA does not exist because matrix A has 2 columns and matrix B has 3 rows. For matrix multiplication to be possible, the number of columns in matrix A must match the number of rows in matrix B.
(d) To find matrix CA where Cϵ R, we need to know the values of matrix C. Since the matrix C is not provided, we cannot compute CA.

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The first elementary row operation that should be performed is D. Replace row 4 with its sum with -3 times row 3. The second crude row operation that should be performed is C. Replace row 3 with its sum with 2 times row 2.

I understand you have provided an accompanying matrix representing a linear system, and you would like to know the first two elementary row operations to perform in solving the system. The matrix you provided appears to be incomplete or not properly formatted. However, I can still guide you on how to approach the problem.

When solving a linear system using an augmented matrix, you would generally perform the following steps:

1. Rearrange the rows, if necessary, so that the pivot (leading entry) in each row is 1 and positioned to the right of the pivot in the row above it.
2. Use row operations to create zeros below the pivots.
3. Use row operations to create zeros above the pivots.
4. Scale each row so that the pivot in each row is 1.

For the first row operation, you can either:
A. Interchange rows to position the pivots correctly, or
B. Scale a row by an integer or a simplified fraction so that the pivot is 1.

For the second row operation, you will most likely replace a row by its sum with a multiple of another row, so that there is a zero below the pivot. Without the correctly formatted matrix, it's difficult to provide a specific answer. However, I hope this general guidance helps you solve the given linear system.

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Here's a proof for each of the statements you provided.

a) If g∘f = I_A, then f is an injection.
Proof: Assume x1 and x2 are elements of A such that f(x1) = f(x2). We want to show that x1 = x2. Since g∘f = I_A, we have g(f(x1)) = g(f(x2)). Applying I_A, we get x1 = g(f(x1)) = g(f(x2)) = x2. Thus, f is injective.

b) If f∘g = I_B, then f is a surjection.
Proof: Let y be an element of B. We want to show that there exists an element x in A such that f(x) = y. Since f∘g = I_B, we have f(g(y)) = I_B(y) = y. Thus, there exists an element x = g(y) in A such that f(x) = y. Therefore, f is surjective.

c) If g∘f = I_A and f∘g = I_B, then f and g are bijections and g = f^(-1).
Proof: From parts (a) and (b), we know that f is both injective and surjective, which means f is a bijection. Similarly, g is also a bijection. Now, we need to show that g = f^(-1). By definition, f^(-1)∘f = I_A and f∘f^(-1) = I_B. Since g∘f = I_A and f∘g = I_B, it follows that g = f^(-1).

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The problem involves finding the area of a shaded region formed by two semicircles drawn on adjacent sides of a square. To solve this problem, we need to find the area of the square and subtract the area of the two semicircles from it.

To find the area of the square, we can simply square the length of its side which is given as 1 unit. So, the area of the square is 1 x 1 = 1 square unit.

Now, to find the area of the shaded region, we need to subtract the area of the two semicircles from the area of the square. The diameter of each semicircle is equal to the length of one of the sides of the square.

Thus, the radius of each semicircle is 1/2 units. Therefore, the area of one semicircle is (π/2) x (1/2)² = π/8 square units. Since there are two semicircles, the total area of the shaded region is (2 x π/8) = π/4 square units. Finally, we can subtract this area from the area of the square to obtain the area of the shaded region which is 1 - π/4 = (4-π)/4 square units.

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Sample size is inversely related to the tolerable deviation rate, a larger sample size is needed to provide a more accurate estimate of the population parameter.

Step-by-step explanation:
1. Sample size refers to the number of observations or units included in a study or analysis to represent a population.
2. Desired level of confidence refers to the degree of certainty that the estimate obtained from the sample accurately represents the population parameter. It is directly related to sample size, as a higher level of confidence generally requires a larger sample.
3. Expected population deviation rate refers to the anticipated rate of deviation or error in a population. It is also directly related to sample size, as a higher expected deviation rate requires a larger sample to ensure accuracy.
4. Tolerable deviation rate, on the other hand, is the maximum rate of deviation that can be accepted in the sample without affecting the overall conclusions. This is inversely related to sample size because as the tolerable deviation rate decreases, a larger sample size is needed to provide a more accurate estimate of the population parameter.

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Answer:

5

Step-by-step explanation:

give me brainliest porfavor

The value of y for the solution set to the given system of equations is :

(e) y = -3

To use Cramer's rule, we need to find the determinant of the coefficient matrix and several other determinants obtained by replacing one column of the coefficient matrix with the constant terms. The coefficient matrix is:

{{-2, 3, -1}, {3, -1, 2}, {-2, 2, -1}}

The determinant of this matrix is:

|-2  3 -1|
| 3 -1  2|
|-2  2 -1| = -12

Now we replace the first column with the constants:

{{-2, 3, -1}, {-1, -1, 2}, {-1, 2, -1}}

The determinant of this matrix is:

|-2  3 -1|
|-1 -1  2|
|-1  2 -1| = 9

Next, we replace the second column with the constants:

{{-2, -2, -1}, {3, -1, 2}, {-2, -1, -1}}

The determinant of this matrix is:

|-2 -2 -1|
| 3 -1  2|
|-2 -1 -1| = 12

Finally, we replace the third column with the constants:

{{-2, 3, -2}, {3, -1, -1}, {-2, 2, -1}}

The determinant of this matrix is:

|-2  3 -2|
| 3 -1 -1|
|-2  2 -1| = -18

Now we can use Cramer's rule to find the value of y. The solution is:

y = D2 / D = 9 / (-12) = -3/4

Therefore, the answer is e) y = -3.

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If we assume that each plant needs one seed to grow, then the number of seeds needed will be equal to the number of plants. The farmer will need 36,000 corn seeds to plant one acre of land.

To find the number of seeds the farmer needs, we first need to determine the area of one acre. One acre is equal to 43,560 square feet. The field shown in the question may have a different area, but we'll assume it's one acre for the purposes of this problem.

Now, we know that the farmer wants to plant 36,000 corn plants per acre. If we assume that each plant needs one seed to grow, then the number of seeds needed will be equal to the number of plants.

Therefore, the farmer will need 36,000 corn seeds to plant one acre of land.we know that the farmer wants to plant 36,000 corn plants per acre.

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Using the intermediate value theorem and Rolle's theorem, we showed that the function [tex]f(x) = ln(x^{2} ) - x + 2[/tex] has exactly one zero on the interval [4,6], which is x = 2.

To show that the function [tex]f(x) = ln(x^{2} ) - x + 2[/tex] has exactly one zero on the interval [4,6], we need to use the intermediate value theorem and Rolle's theorem.

First, we can find that the function is continuous and differentiable for x > 0. Taking the derivative of f(x), we get [tex]f'(x) = (2/x) - 1[/tex]. Setting f'(x) = 0, we get x = 2.

Now, let's evaluate f(4) and f(6). We have [tex]f(4) = ln(16) - 4 + 2 = ln(16) - 2[/tex]and [tex]f(6) = ln(36) - 6 + 2 = ln(36) - 4[/tex]. Using a calculator, we find that f(4) < 0 and f(6) > 0.

By the intermediate value theorem, since f(x) is continuous on [4,6] and takes on values of opposite signs at the endpoints, there exists at least one zero of f(x) on the interval.

Finally, to show that there is only one zero, we use Rolle's theorem. Since f(x) is differentiable on (4,6) and has a zero on this interval, there must exist at least one point c in (4,6) such that f'(c) = 0.

From earlier, we know that f'(x) = (2/x) - 1, so we have [tex]f'(c) = (2/c) - 1 = 0[/tex], which implies c = 2. Therefore, the only zero of f(x) on [4,6] is x = 2.

In summary, using the intermediate value theorem and Rolle's theorem, we showed that the function [tex]f(x) = ln(x^{2} ) - x + 2[/tex] has exactly one zero on the interval [4,6], which is x = 2.

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Answer:

x = 14

Step-by-step explanation:

using the cosine ratio in the right triangle and the exact value

cos30° = [tex]\frac{\sqrt{3} }{2}[/tex] , then

cos30° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{7\sqrt{3} }{x}[/tex] = [tex]\frac{\sqrt{3} }{2}[/tex] ( cross- multiply )

x × [tex]\sqrt{3}[/tex] = 14[tex]\sqrt{3}[/tex] ( divide both sides by [tex]\sqrt{3}[/tex] )

x = 14

Each piece of wood that Ren cuts will be 3/8 of a meter long.

To solve the problem, we need to divide 3/4 by 2. This can be written as:

3/4 ÷ 2

To model this using fraction bars, we can start by drawing a bar to represent the whole piece of wood, which is 3/4 of a meter long:

___________________

|___________________|

        3/4

Next, we need to divide this bar into 2 equal parts. We can do this by drawing a line down the middle of the bar:

_______ _______

|_______|_______|

  3/4     3/4

Now we can see that we have two equal pieces of wood, each of which is 3/4 ÷ 2 = 3/8 of a meter long.

To calculate this, we can divide the numerator (3) by 2 to get 1.5, and then write this as a fraction with a denominator of 8:

1.5 ÷ 2 = 0.75

0.75 = 3/4

3/4 ÷ 2 = 3/8

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To determine the probability of a student getting exactly 10 questions correct, we need to know the total number of ways in which the student can answer the questions and the number of ways in which the student can get exactly 10 questions correct.

Assuming that each question has only two possible answers (e.g. true/false or multiple choice with two options), and the student guesses randomly, the probability of getting a single question correct is 1/2, and the probability of getting a single question incorrect is also 1/2.

Let X be the number of questions the student answers correctly, and n be the total number of questions.

In this case, n = 20 (the total number of questions), and p = 1/2 (the probability of getting a single question correct).

where (n choose k) is the binomial coefficient, which represents the number of ways to choose k items from a set of n items, and is given by:

(n choose k) = n! / (k! * (n - k)!)

P(X = 10) = (20 choose 10) * (1/2)^10 * (1/2)^(20-10)

= 184,756 * 0.0009765625 * 0.0009765625

= 0.1801

Therefore, the probability of the student getting exactly 10 questions correct is 0.1801, or approximately 18.01%.

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Of the joint probability function

(a) The value of the constant c is approximately 0.0238.

(b) P(X=0,Y=3) ≈ 0.0714.

(c) P(X≥ 0,Y≤ 1) ≈ 0.4524.

(d) The given statement "X and Y are not independent" is False.

(a) To find the value of the constant c, we need to use the fact that the sum of the probabilities over all possible values of X and Y must be equal to 1:

∑∑f(x,y) = 1

∑x=[tex]0^2[/tex] ∑y=[tex]0^3[/tex] c(2x+y) = 1

c(0+1+2+3+2+3+4+5+4+5+6+7) = 1

c(42) = 1

c = 1/42 ≈ 0.0238 (rounded to three decimal places)

(b) P(X=0,Y=3) = f(0,3) = c(2(0)+3) = 3c = 3(1/42) ≈ 0.0714 (rounded to three decimal places)

(c) P(X≥0,Y≤1) = f(0,0) + f(0,1) + f(1,0) + f(1,1) + f(2,0) + f(2,1)

= c(2(0)+0) + c(2(0)+1) + c(2(1)+0) + c(2(1)+1) + c(2(2)+0) + c(2(2)+1)

= c(1+3+2+4+4+5) = 19c = 19(1/42) ≈ 0.4524 (rounded to three decimal places)

(d) We can check whether X and Y are independent by verifying if P(X=x,Y=y) = P(X=x)P(Y=y) for all possible values of X and Y. Let's check this for some cases:

P(X=0,Y=0) = f(0,0) = c(2(0)+0) = 0

P(X=0) = f(0,0) + f(0,1) + f(0,2) + f(0,3) = c(0+1+2+3) = 6c

P(Y=0) = f(0,0) + f(1,0) + f(2,0) = c(0+2+4) = 6c

P(X=0)P(Y=0) = [tex]36c^2[/tex]

Since P(X=0,Y=0) ≠ P(X=0)P(Y=0), X and Y are not independent. Therefore, the answer is (C) false.

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The volume of the prism is determined as 63 cm³.

The volume of the triangular prism is calculated by applying the following formula as shown below;

V = ¹/₂bhl

where;

The volume of the prism is calculated as follows;

V = ¹/₂ x 7 cm x 3 cm x 6 cm

V = 63 cm³

,

Thus, the volume of the prism is a function of its base, height and length.

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Since the difference is negative (-8), the sequence is monotonic decreasing.

A monotonic function in mathematics is a function between ordered sets that maintains or flips the given order. Calculus was where this idea initially surfaced, and it was later applied to the more abstract context of order theory.  If the variables Yj can be arranged so that if Yj is missing, then all variables Yk with k>j are likewise missing, then the pattern of missing data is said to be monotone.

This happens, for instance, in drop-out-prone longitudinal research. The pattern is said to as non-monotone or generic if it is not monotonous.

Using the difference test to calculate the nth term of the sequence, we get:

[tex]a_n - a_{n-1} \\= 6 - 8n - (6 - 8(n-1)) \\= -8[/tex]

Since the difference is negative (-8), the sequence is monotone decreasing.

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To compute the directional Gervative of the function f(x,y) = exy - x - 2y at the point P = (2,-1) in the direction of the vector v = , we first need to find the gradient of f at P.

The gradient of f is given by ∇f(x,y) = . So, at the point P = (2,-1), we have ∇f(2,-1) = .

Next, we need to find the unit vector in the direction of v. To do this, we first need to find the magnitude of v, which is ||v|| = √(e^2 + (-2)^2) = √(e^2 + 4).

Then, we can find the unit vector in the direction of v by dividing v by its magnitude:

u = v/||v|| = .

Finally, we can compute the directional Gervative of f at P in the direction of v as follows:

D_v f(2,-1) = ∇f(2,-1) · u = ( · )

= (e^-1 - 1)(e/√(e^2 + 4)) + (e^2 - 2)(-2/√(e^2 + 4))

= -2e/(√(e^2 + 4)) - 4/(√(e^2 + 4))

= (-2e - 4)/(√(e^2 + 4)).

Therefore, the directional Gervative of f at P in the direction of v is (-2e - 4)/(√(e^2 + 4)).

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Check the picture below, so the parabola looks more or less like so, with a positive "p" distance of 2, with the vertex half-way between the directrix and the focus point.

[tex]\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{p~is~negative}{op ens~\supset}\qquad \stackrel{p~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\begin{cases} h=8\\ k=7\\ p=2 \end{cases}\implies 4(2)(~~x-8~~) = (~~y-7~~)^2 \implies 8(x-8)=(y-7)^2 \\\\\\ x-8=\cfrac{1}{8}(y-7)^2\implies {\Large \begin{array}{llll} x=\cfrac{1}{8}(y-7)^2+8 \end{array}}[/tex]