What 11/5 cubic feet = to?

Expand to write an equivalent expression <br> -1/4 (-8x + 12y)

Tanzania [10]

The equivalent expression on expanding -1/4 (-8x + 12y) is 2x - 3y

<h3>Expansion of expression</h3>

Expansion of expression can be carried out using the distributive rule a shown:

A(B + C) = AB + AC

Note that <u>A is distributed over B and C</u>
Given the expression

-1/4 (-8x + 12y)

-1/4(-8x) - 1/4(12y)

2x - 3y

Hence the equivalent expression on expanding -1/4 (-8x + 12y) is 2x - 3y

Learn more on expansion here: brainly.com/question/26430239

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7 0

1 year ago

According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe

Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .

In this problem:

22% of couples meet online, hence p = 0.22.

A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

$\mu = p = 0.22$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338$

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

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

By the Central Limit Theorem:

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

$Z = \frac{0.25 - 0.22}{0.0338}$

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

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

$Z = \frac{0.2 - 0.22}{0.0338}$

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

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

$Z = \frac{0.15 - 0.22}{0.0338}$

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

4 0

1 year ago

What is the domain and range of this function?

max2010maxim [7]

As for the domain, the only restriction comes from the logarithm. The argument of a logarith must be strictly positive, so we have

$x-2>0 \iff x>2$

As for the range, we have:

The range of $\log(x)$ are all real numbers
If we change to $\log(x-2)$ we're translating the function horizontally, so the range remains the same
If we change to $\log(5(x-2()$ we're stretching the function horizontally, so the range doesn't change
If we change to $\log(5(x-2))+1$ we're translating the function 1 unit up, but the range is already all the real numbers, so it doesn't change.