1. Which of the following is not a true statement?

To find the volume of a cube with sides of ¼ you……..

const2013 [10]

Answer:

1/64

Step-by-step explanation:

Volume of cube = length^3 = (1/4)^3= 1/64

7 0

3 years ago

Solve the question below, please.

Ne4ueva [31]

Answer:

Adam: 52.26°

Step-by-step explanation:

Use Cosine Rule, $Cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$ , to find the angle of both players.

✔️Angle of Carlos:

Let the angle be C,

a = 50 ft

b = 40 ft

c = 24 ft

Plug in the values into the equation

$Cos(C) = \frac{50^2 + 40^2 - 24^2}{2*50*40}$

$Cos(C) = \frac{3,524}{4,000}$

$C = Cos^{-1}(\frac{3,524}{4,000})$

C = 28.24° (nearest hundredth)

✔️Angle of Adam:

Let the angle be C,

a = 30 ft

b = 22 ft

c = 24 ft

Plug in the values into the equation

$Cos(C) = \frac{30^2 + 22^2 - 24^2}{2*30*22}$

$Cos(C) = \frac{808}{1,320}$

$C = Cos^{-1}(\frac{808}{1,320)}$

C = 52.26° (nearest hundredth)

6 0

3 years ago

Eight children were in a ballet class. Three of them were wearing a pink leotard. What fraction of them did not have a pink leot

leva [86]

Answer:

5/8

Step-by-step explanation:

8 children in class

3 have pink on

8-3 = 5

That means 5 do not have pink

Fraction that do not have pink

those that do not have pink/total

5/8

5 0

3 years ago

Becky is planning a 2 100-mile trip to St. Louis to visit a college. Her car averages 30 miles per gallon

GaryK [48]

Answer:

70

Step-by-step explanation:

3 0

3 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3

In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that $\mu = 8.3, \sigma = 3.3$ .

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

$s = \frac{3.3}{\sqrt{37}} = 0.5425$

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

This probability is the pvalue of Z when $X = 8.65$ . So

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

$Z = \frac{8.65 - 8.3}{0.5425}$

$Z = 0.65$

$Z = 0.65$ has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is subtracted by the pvalue of Z when $X = 7.43$

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

$Z = \frac{7.43 - 8.3}{0.5425}$

$Z = -1.60$

$Z = -1.60$ has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is the pvalue of Z when $X = 8.65$ subtracted by the pvalue of Z when $X = 7.43$ .

So:

From a), we have that for $X = 8.65$ , we have $Z = 0.65$ , that has a pvalue of 0.7422.

From b), we have that for $X = 7.43$ , we have $Z = -1.60$ , that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0

3 years ago