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1 year ago

Find the quotient. Simplify your answer.

Mathematics

1 answer:

brilliants [131]1 year ago

8 0

Let's see

$\\ \rm\Rrightarrow \dfrac{s+1}{s^2+1}\div \dfrac{s^2}{6s^2}$

$\\ \rm\Rrightarrow \dfrac{s+1}{s(s+1)}\div\dfrac{1}{5s^2}$

$\\ \rm\Rrightarrow \dfrac{1}{s}\times{5s^2}$

$\\ \rm\Rrightarrow \dfrac{5s^2}{s}$

$\\ \rm\Rrightarrow 5s$

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Two groups of students were asked how far they lived from their school. The table below shows the distances in miles:

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Answer:

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2 years ago

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Suppose that the sample standard deviation was s = 5.1. Compute a 98% confidence interval for μ, the mean time spent volunteerin

NISA [10]

Answer:

The 95% confidence interval would be given by (5.139;5.861)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

2) Confidence interval

Assuming that $\bar X =5.5$ and the ranfom sample n=1086.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=1086-1=1085$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,1085)".And we see that $t_{\alpha/2}=2.33$

Now we have everything in order to replace into formula (1):

$5.5-2.33\frac{5.1}{\sqrt{1086}}=5.139$

$5.5+2.333\frac{5.1}{\sqrt{1086}}=5.861$

So on this case the 95% confidence interval would be given by (5.139;5.861) We are 98% confident that the mean time spent volunteering for the population of parents of school-aged children is between these two values.

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