using the factor theorem, which polynomial function has the zeros 4 and 4 – 5i? x3 – 4x2 – 23x 36 x3 – 12x2 73x – 164 x2 – 8x – 5ix 20i 16 x2 – 5ix – 20i – 16

The Value of B[ex-2Y+Z] is e^(-7/2) - 1/5 + 2.

To compute B[ex-2Y+Z], we need to determine the probability distribution of the expression ex-2Y+Z.

Given that X ~ N(-4,1), Y ~ Exp(10), and Z ~ Poisson(2) are independent, we can start by calculating the mean and variance of each random variable:

For X ~ N(-4,1):

Mean (μ) = -4

Variance (σ^2) = 1

For Y ~ Exp(10):

Mean (μ) = 1/λ = 1/10

Variance (σ^2) = 1/λ^2 = 1/10^2 = 1/100

For Z ~ Poisson(2):

Mean (μ) = λ = 2

Variance (σ^2) = λ = 2

Now let's calculate the expression ex-2Y+Z:

B[ex-2Y+Z] = E[ex-2Y+Z]

Since X, Y, and Z are independent, we can calculate the expected value of each term separately:

E[ex] = e^(μ+σ^2/2) = e^(-4+1/2) = e^(-7/2)

E[2Y] = 2E[Y] = 2 * (1/10) = 1/5

E[Z] = λ = 2

Now we can substitute these values into the expression:

B[ex-2Y+Z] = E[ex-2Y+Z] = e^(-7/2) - 1/5 + 2

Therefore, the value of B[ex-2Y+Z] is e^(-7/2) - 1/5 + 2.

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The coefficient of determination is often used to evaluate the usefulness of regression models.

In simple linear regression, r2 is the coefficient of determination. In statistics, a measure of the proportion of the variance in one variable that can be explained by another variable is referred to as the coefficient of determination (R2 or r2).

The coefficient of determination, often known as the squared correlation coefficient, is a numerical value that indicates how well one variable can be predicted from another using a linear equation (regression).The coefficient of determination is always between 0 and 1, with a value of 1 indicating that 100% of the variability in one variable is due to the linear relationship between the two variables in question.

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The missing value to create a probability distribution is 0.46.

To find the missing value required to create a probability distribution, we need to add the probabilities and subtract from 1.

This is because the sum of all the probabilities in a probability distribution must be equal to 1.

Here is the given probability distribution:x / P(x)0 / 0.181 / 0.112 / 0.133 / 4 / 0.12

Let's add up the probabilities:

0.18 + 0.11 + 0.13 + 0.12 + P(4) = 1

Simplifying, we get:0.54 + P(4) = 1

Subtracting 0.54 from both sides, we get

:P(4) = 1 - 0.54P(4)

= 0.46

Therefore, the missing value to create a probability distribution is 0.46.

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Mr. Potatohead will be carried downstream by 10 × 43.5 = 435 meters approximately.

Given, Velocity of water (vw) = 10 m/s Velocity of Mr. Potatohead (vp) = 1.2 m/s

Distance between Mr. Potatohead and the waterfall (d) = 40 m Angle (θ) = 40

The velocity of Mr. Potatohead with respect to ground can be calculated by using the Pythagorean theorem.

Using this theorem we can find the horizontal and vertical components of the velocity of Mr. Potatohead with respect to ground.

vp = (vpx2 + vpy2)1/2 ......(1)

The horizontal and vertical components of the velocity of Mr. Potatohead with respect to ground are given as,

vpx = vp cos θ

vpy = vp sin θ

On substituting these values in equation (1),

vp = [vp2 cos2θ + vp2 sin2θ]1/2

vp = vp [cos2θ + sin2θ] 1/2

vp = vp

Therefore, the velocity of Mr. Potatohead with respect to the ground is 1.2 m/s.

Since Mr. Potatohead is swimming at an angle of 40°, the horizontal component of his velocity with respect to the ground is,

vpx = vp cos θ

vpx = 1.2 cos 40°

vpx = 0.92 m/s

As per the question, Mr. Potatohead is attempting to cross a river flowing at 10 m/s from a point 40 m away from a treacherous waterfall.

To find how far Mr. Potatohead is carried downstream, we can use the equation, d = vw t,

Where, d = distance carried downstream vw = velocity of water = 10 m/sand t is the time taken by Mr. Potatohead to cross the river.

The time taken by Mr. Potatohead to cross the river can be calculated as, t = d / vpx

Substituting the values of d and vpx in the above equation,

we get t = 40 / 0.92t

≈ 43.5 seconds

Therefore, Mr. Potatohead will be carried downstream by 10 × 43.5 = 435 meters approximately.

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