A triangle is classified as an equilateral triangle when it has

The oblique pyramid has a square base with an edge length of 5 cm. The height of the pyramid is 7 cm. What is the volume of the

ExtremeBDS [4]

The answer is <span>58 cm3
</span>
The volume of the pyramid with square base is:
V = s² * h / 3
s - the edge length, s = 5 cm
h - height, h = 7 cm

V = 5² * 7 / 3
V = 25 * 7 / 3
V = 58 cm³

7 0

3 years ago

The average amount of water in randomly selected 16-ounce bottles of water is 16.15 ounces with a standard deviation of 0.45 oun

vekshin1

Answer:

0.0179 = 1.79% probability that the mean of this sample is less than 15.99 ounces of water.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes 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 $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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

In this question, we have that:

$\mu = 16.15, \sigma = 0.45, n = 35, s = \frac{0.45}{\sqrt{35}} = 0.0761$

What is the probability that the mean of this sample is less than 15.99 ounces of water?

This is the pvalue of Z when X = 15.99. So

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

By the Central Limit Theorem

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

$Z = \frac{15.99 - 16.15}{0.0761}$

$Z = -2.1$

$Z = -2.1$ has a pvalue of 0.0179

0.0179 = 1.79% probability that the mean of this sample is less than 15.99 ounces of water.

3 0

3 years ago

Omg please help please

Brut [27]

$x^{2}$ Answer:

(Explanation)

Step-by-step explanation:

Part A:

The graph of y = $x^{2}$ + 2 will be translated 2 units up from the graph of y = $x^{2}$ .

If you plug in 0 for x, you get a y-value of 2. The 2 is also not included with the $x^{2}$ , which is why it doesn't translate left.

This is what graph A should look like:

[Attached File]

Part B:

The graph of y = $x^{2}$ - 2 will be translated 2 units down from the graph of y = $x^{2}$ .

If you plug in 0 for x, you get a y-value of -2. The 2 is also not included with the $x^{2}$ , which is why it doesn't translate right.

This is what graph B should look like:

[Attached File]

Part C:

The graph of y = 2 $x^{2}$ is a stretched version of the graph y = $x^{2}$ . Numbers that are greater than 1 stretch and open up and numbers less than -1 stretch and open down.

This is what graph C should look like:

[Attached File]

Part D:

The graph of y = $\frac{1}{2}x^{2}$ is a compressed version of the graph y = $x^{2}$ . Numbers that are in-between 0 and 1, and -1 and 0 are compressed.