Question 6 of 10

Write an equation for the inverse function y = -22x -7

Fed [463]

Answer: $\frac{-7-y}{22}$

Step-by-step explanation:

7 0

3 years ago

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Help me pleasee , timed

solong [7]

Answer:

The second alternative is correct

Step-by-step explanation:

We have been given the expression;

$(x^{27}y)^{\frac{1}{3}}$

The above expression can be re-written as;

$(x^{27})^{\frac{1}{3}}*y^{\frac{1}{3}}\\\\(x^{27})^{\frac{1}{3}}=x^{27*\frac{1}{3}}=x^{9}$

On the other hand;

$y^{\frac{1}{3}}=\sqrt[3]{y}$

Therefore, we have;

$x^{9}\sqrt[3]{y}$

6 0

3 years ago

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Juries should have the same racial distribution as the surrounding communities. According to the U.S. Census Bureau, 18% of resi

Ymorist [56]

Answer:

0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

18% of residents in Minneapolis, Minnesota, are African Americans. Suppose a local court will randomly sample 100 state residents and will then observe the proportion in the sample who are African American.

This means that $p = 0.18, n = 100$