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Andrew [12]
2 years ago
13

2. Which algebraic expression represents the following phrase?

Mathematics
1 answer:
oksian1 [2.3K]2 years ago
3 0

The algebraic expression which represents the square of the difference of s and 6 is (s - 6)²

Step-by-step explanation:

Let us represent some words by mathematics expressions

  • Square a number ⇒ x²
  • Square the sum of two numbers ⇒ (x + y)²
  • Sum of the squares of two numbers ⇒ x² + y²
  • Square of difference of two numbers (x - y)²
  • The difference of square two numbers x² - y²

∵ The expression is the square of the difference of s and 6

- That means find the difference between s and 6 at first,

   then square the difference

∵ The difference of s and 6 = s - 6

- Square this difference means but the difference in a bracket

   and then square the bracket

∴ The square of the difference = (s - 6)²

The algebraic expression which represents the square of the difference of s and 6 is (s - 6)²

Learn more:

You can learn more about the algebraic expressions in brainly.com/question/10771256

#LearnwithBrainly

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Explained below.

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The standard deviation of this sampling distribution of sample proportion is:

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Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0204^{2}).

(b)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

P(p-0.04

                                          =P(-1.96

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.95.

(c)

The sample selected is of size <em>n</em> = 200 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{200}}=0.0306

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0306^{2}).

(d)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

P(p-0.04

                                          =P(-1.31

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.81.

(e)

The probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 450 is 0.95.

And the probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 200 is 0.81.

So, there is a gain in precision on increasing the sample size.

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2 years ago
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