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Rzqust [24]
2 years ago
11

Multiply 7x(x^- x+2)

Mathematics
1 answer:
Pie2 years ago
6 0

Rewrite (7x−2) 2 (7x-2) 2 as (7x−2) (7x−2) (7x-2) (7x-2).

(7x−2) (7x−2) (7x-2) (7x-2)

Expand (7x−2)(7x−2)(7x-2)(7x-2) using the FOIL Method.

Apply the distributive property.

7x (7x−2) −2 (7x−2) 7x (7x-2)-2 (7x-2)

Apply the distributive property.

7x(7x)+7x⋅−2−2(7x−2)7x (7x)+7x⋅-2-2(7x-2)

Apply the distributive property.

7x(7x)+7x⋅−2−2(7x)−2⋅−27x(7x)+7x⋅-2-2(7x)-2⋅-2

Simplify and combine like terms.

Simplify each term.

Multiply xx by xx.

7⋅7x2+7x⋅−2−2(7x)−2⋅−27⋅7x2+7x⋅-2-2(7x)-2⋅-2

Multiply 77 by 77.

49x2+7x⋅−2−2(7x)−2⋅−249x2+7x⋅-2-2(7x)-2⋅-2

Multiply −2-2 by 77.

49x2−14x−2(7x)−2⋅−249x2-14x-2(7x)-2⋅-2

Multiply 77 by −2-2.

49x2−14x−14x−2⋅−249x2-14x-14x-2⋅-2

Multiply −2-2 by −2-2.

49x2−14x−14x+449x2-14x-14x+4

Subtract 14x14x from −14x-14x.

49x2−28x+4

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Simplify, and write your answer as a numeral:<br><br> ( The problem is in the picture I attached )
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The average Act score follows a normal distribution, with a mean of 21.1 and a standard deviation of 5.1. What is the probabilit
ohaa [14]

Answer:

0.43% probability that the mean IQ score of 50 randomly selected people will be more than 23

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are solved using 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:

\mu = 21.1, \sigma = 5.1, n = 50, s = \frac{5.1}{\sqrt{50}} = 0.7212

What is the probability that the mean IQ score of 50 randomly selected people will be more than 23

This is 1 subtracted by the pvalue of Z when X = 23. So

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

By the Central Limit Theorem

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

Z = \frac{23 - 21.1}{0.7212}

Z = 2.63

Z = 2.63 has a pvalue of 0.9957

1 - 0.9957 = 0.0043

0.43% probability that the mean IQ score of 50 randomly selected people will be more than 23

5 0
3 years ago
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