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3 years ago

48 ÷ [(5 + 16) − 13]

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

3 0

Answer:

Step-by-step explanation:

48 ÷ [(5 + 16) − 13]

PEMDAS work from the inside out

48 ÷ [21 − 13]

48 ÷ [8]

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Solve, using the substitution method.

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14-4x=2y
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subsitute 7-2x for y
7-2x+2x=7
add like terms
7=7
true
therefor these 2 equations are the same and there are an infinite number of equations

answer is C

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Randall bought 2 game controllers at Electronics Plus for $36. What is the unit rate for a game controller at Electronics Plus

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In 2008 the Better Business Bureau settled 75% of complaints they received (USA Today, March 2, 2009). Suppose you have been hir

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Answer:

Explained below.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}= p$

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

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

(a)

The sample selected is of size <em>n</em> = 450 > 30.

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

The mean and standard deviation are:

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

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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Answer:

Step-by-step explanation:

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