What is [(x+4)^2]^3 simplified

The average production cost for major movies is 57 million dollars and the standard deviation is 22 million dollars. Assume the

Degger [83]

Using the normal distribution, we have that:

The distribution of X is $X \approx (57,22)$ .

The distribution of $\mathbf{\bar{X}}$ is $\bar{X} \approx (57, 5.3358)$ .

0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem, the parameters are given as follows:

$\mu = 57, \sigma = 22, n = 17, s = \frac{22}{\sqrt{17}} = 5.3358$

Hence:

The distribution of X is $X \approx (57,22)$ .

The distribution of $\mathbf{\bar{X}}$ is $\bar{X} \approx (57, 5.3358)$ .

The probabilities are the <u>p-value of Z when X = 58 subtracted by the p-value of Z when X = 55</u>, hence, for a single movie:

X = 58:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{58 - 57}{22}$

Z = 0.05.

Z = 0.05 has a p-value of 0.5199.

X = 55:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55 - 57}{22}$

Z = -0.1.

Z = -0.1 has a p-value of 0.4602.

0.5199 - 0.4602 = 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

For the sample of 17 movies, we have that:

X = 58:

$Z = \frac{X - \mu}{s}$

$Z = \frac{58 - 57}{5.3358}$

Z = 0.19.

Z = 0.19 has a p-value of 0.5753.

X = 55:

$Z = \frac{X - \mu}{s}$

$Z = \frac{55 - 57}{5.3358}$

Z = -0.38.

Z = -0.38 has a p-value of 0.3520.

0.5753 - 0.3520 = 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

More can be learned about the normal distribution at brainly.com/question/4079902

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8 0

2 years ago

A hardware store sells light bulbs in different quantities. the graph shows the cost of various quantities. according to the gra

skelet666 [1.2K]

Looking at the two black dots

5 bulbs cost $9

Divide total cost by number of bulbs bought:

9 / 5 = $1.80 per bulb.

10 bulbs cost $18

Divide total cost by number of bulbs bought:

18 / 10 = $1.80 per bulb

The cost for one bulb is $1.80

3 0

4 years ago

Is RS perpendicular to DF? Select Yes or No for each statement. R (6, −2), S (−1, 8), D (−1, 11), and F (11 ,4) R (1, 3), S (4,7

guajiro [1.7K]

I'll do the first one to get you started.

Find the slope of the line between R (6,-2) and S (-1,8) to get

m = (y2-y1)/(x2-x1)

m = (8-(-2))/(-1-6)

m = (8+2)/(-1-6)

m = 10/(-7)

m = -10/7

The slope of line RS is -10/7

Next, we find the slope of line DF

m = (y2 - y1)/(x2 - x1)

m = (4-11)/(11-(-1))

m = (4-11)/(11+1)

m = -7/12

From here, we multiply the two slope values

(slope of RS)*(slope of DF) = (-10/7)*(-7/12)

(slope of RS)*(slope of DF) = (-10*(-7))/(7*12)

(slope of RS)*(slope of DF) = 10/12

(slope of RS)*(slope of DF) = 5/6

Because the result is not -1, this means we do not have perpendicular lines here. Any pair of perpendicular lines always has their slopes multiply to -1. This is assuming neither line is vertical.

I'll let you do the two other ones. Let me know what you get so I can check your work.

3 0

3 years ago

|2x + 1| - 4 < 5 <br><br> solve and graph

NeX [460]

Answer:

The value of x could be 3

Step-by-step explanation:

2 x 3 = 6 + 1 - 4 = 3 which is less than 5

7 0

3 years ago

Read 2 more answers

The arch beneath a bridge is semi-elliptical, a one-way roadway passes under the arch. The width of the roadway is 38 feet and

forsale [732]

Answer:

Only truck 1 can pass under the bridge.

Step-by-step explanation:

So, first of all, we must do a drawing of what the situation looks like (see attached picture).

Next, we can take the general equation of an ellipse that is centered at the origin, which is the following:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}$

where:

a= wider side of the ellipse

b= shorter side of the ellipse

in this case:

$a=\frac{38}{2}=19ft$

and

b=12ft

so we can go ahead and plug this data into the ellipse formula:

$\frac{x^2}{(19)^2}+\frac{y^2}{(12)^2}$

and we can simplify the equation, so we get:

$\frac{x^2}{361}+\frac{y^2}{144}$

So, we need to know if either truk will pass under the bridge, so we will match the center of the bridge with the center of each truck and see if the height of the bridge is enough for either to pass.

in order to do this let's solve the equation for y:

$\frac{y^{2}}{144}=1-\frac{x^{2}}{361}$

$y^{2}=144(1-\frac{x^{2}}{361})$

we can add everything inside parenthesis so we get:

$y^{2}=144(\frac{361-x^{2}}{361})$

and take the square root on both sides, so we get:

$y=\sqrt{144(\frac{361-x^{2}}{361})}$

and we can simplify this so we get:

$y=\frac{12}{19}\sqrt{361-x^{2}}$

and now we can evaluate this equation for x=4 (half the width of the trucks) so:

$y=\frac{12}{19}\sqrt{361-(8)^{2}}$

y=11.73ft

this means that for the trucks to pass under the bridge they must have a maximum height of 11.73ft, therefore only truck 1 is able to pass under the bridge since truck 2 is too high.

5 0

3 years ago