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Fynjy0 [20]
3 years ago
6

Choose any four consecutive even numbers. (For example: 6, 8, 10, 12). Multiply the two middle numbers together. (e.g. 8 x 10 =

80) Multiply the first and last numbers. (e.g. 6 x 12 = 72) Now subtract your second answer from the first. (e.g. 80 - 72 = 8) Try it with your own numbers. Why is the answer always 8?
Mathematics
1 answer:
Ludmilka [50]3 years ago
6 0

Answer:

Step-by-step explanation:

let the consecutive even numbers be  2n-2,2n,2n+2,2n+4 where n>0 and n is an integer.

2n(2n+2)-(2n-2)(2n+4)

=4n²+4n-(4n²+8n-4n-8)

=4n²+4n-4n²-8n+4n+8

=8

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Step-by-step explanatU

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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
Ede4ka [16]

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:

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

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(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

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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

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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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I don’t know what the last one is
joja [24]

But its still not that much info what is the answers for it if there is no answer

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