What are the solutions to the equation

The volume of a cube whose edge is 3x, is

diamong [38]

Answer:

$27 { x}^{3}$

Step-by-step explanation:

s represents one side of the cube

Volume of cube= (s)^3

= (3x)^3

= 27x^3

6 0

3 years ago

Please help me. Please put correct answer

kicyunya [14]

Mixed Number Form:

$- 1 \frac1 2$

5 0

3 years ago

Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72

Lisa [10]

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that $\mu = 38.72, \sigma = 3.17$

Sample of 10:

This means that $n = 10, s = \frac{3.17}{\sqrt{10}}$

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

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

By the Central Limit Theorem

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

$Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}$

$Z = 1.28$

$Z = 1.28$ has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

$\mu = 266, \sigma = 16$

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

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

$Z = \frac{260 - 266}{16}$

$Z = -0.375$

$Z = -0.375$ has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now $n = 20$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}$

$Z = -1.68$

$Z = -1.68$ has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now $n = 50$ , so:

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

$Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}$

$Z = -2.65$

$Z = -2.65$ has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that $n = 15$ . This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

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

$Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}$

$Z = 2.42$

$Z = 2.42$ has a p-value of 0.9922.

X = 256

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

$Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}$

$Z = -2.42$

$Z = -2.42$ has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

8 0

3 years ago

A company manufactures keyboards for desktop and laptop computers. The desktop keyboard contains 105 keys and the laptop keyboar

Luba_88 [7]

The company can produce 20 desktop keyboards and 10 laptop keyboards

Step-by-step explanation:

A company manufactures keyboards for desktop and laptop computers

The desktop keyboard contains 105 keys
The laptop keyboard has 90 keys
The company has 3000 keys
The company needs to produce twice as many desktop keyboards as laptop keyboards

We need to find how much of each type the company can produce

Assume that the company can produce x desktop keyboards and

y laptop keyboards

∵ The company can produce x desktop keyboards

∵ company can produce y laptop keyboards

∵ The desktop keyboard contains 105 keys

∵ The laptop keyboard has 90 keys

∵ The company has 3000 keys

- Write an equation to represent the total keys

∴ 105 x + 90 y = 3000 ⇒ (1)

∵ The company needs to produce twice as many desktop keyboards

as laptop keyboards

∴ x = 2 y ⇒ (2)

Substitute x in equation (1) by equation (2)

∵ 105(2 y) + 90 y = 3000

∴ 210 y + 90 y = 3000

∴ 300 y = 3000

- Divide both sides by 300

∴ y = 10

∴ The company can produce 10 laptop keyboards

Substitute the value of y in equation (2) to find the value of x

∵ x = 2(10)

∴ x = 20

∴ The company can produce 20 desktop keyboards

The company can produce 20 desktop keyboards and 10 laptop keyboards

Learn more:

You can learn more about the system of equations in brainly.com/question/2115716

#LearnwithBrainly

6 0

3 years ago

Planes X and Y and points C, D, E, and F are shown.

viva [34]

Answer:

The answer is below

Step-by-step explanation:

The planes are shown in the image below.

From the image, points D and E are on plane Y, while point F is on plane X.

A) The line that can be drawn through points C and D is contained in plane Y.

This statement is wrong because point C is not in plane Y

B) The line that can be drawn through points D and E is contained in plane Y.

Correct. Since both points D and E are on plane Y, hence The line that can be drawn through points D and E is contained in plane Y.

C) The only point that can lie in plane X is point F.

This statement is correct because from the image only point F is on plane X.

D) The only points that can lie in plane Y are points D and E.

This statement is correct because from the image only point D and E is on plane Y.

5 0

3 years ago