Use the distributive property to rewrite the expression –2(6

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babunello [35]

K should be ur answer

3 0

2 years ago

(3t² + 2 + -2) + (t² +3t - 4) - (4 + -5)

enot [183]

(3t^2 + 2 + (-2)) + (t^2 + 3t - 4) - (4 + (-5))

Apply the distributive property to remove parentheses.

3t^2 + 2 - 2 + t^2 + 3t - 4 - 4 + 5

Combine like terms.

<h3>4t^2 + 3t - 3 is the simplified form of the given expression.</h3>

8 0

3 years ago

Find the nonpermissible replacement for b in b - 1/ b + 7

Juliette [100K]

This will be -7 because this would make the denominator b + 7 = 0.

3 0

3 years ago

Which system of equations could be graphed to solve the equation below? log Subscript 0.5 Baseline x = log Subscript 3 Baseline

mariarad [96]

Answer:

Step-by-step explanation:

y 1 = StartFraction log x Over log 0.5 EndFraction, y 2 = StartFraction log 2 Over log 3 EndFraction + x

3 0

3 years ago

A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the

Orlov [11]

Answer:

95.44% probability that the sample proportion will differ from the population proportion by less than 0.03.

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 $s = \sqrt{\frac{p(1-p)}{n}}$