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4 years ago

A grid shows the positions of a subway stop and your house. The subway stop is located at (7, -6) and your house is located at (

-9, 5). What is the distance, to the nearest unit, between your house and the subway stop?

Mathematics

1 answer:

Westkost [7]4 years ago

5 0

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 7}}\quad ,&{{ -6}})\quad 
% (c,d)
&({{ -9}}\quad ,&{{ 5}})
\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
d=\sqrt{[-9-7]^2+[5-(-6)]^2}\implies d=\sqrt{[-9-7]^2+[5+6]^2}$

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ziro4ka [17]

Answer:

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Step-by-step explanation:

(1,-8)

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3 years ago

As part of their business promotional package, the Milwaukee Chamber of Commerce would like an estimate of the mean cost per mon

Morgarella [4.7K]

Answer:

a) $884-2.426\frac{50}{\sqrt{40}}=864.82$

$884+2.426\frac{50}{\sqrt{40}}=903.179$

So on this case the 98% confidence interval would be given by (864.82;903.179)

b) For this case the upper bound of the confidence interval is lower than 950 so we don't have a significant evidence to conclude that the true mean is 950 per month

Step-by-step explanation:

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".

$\bar X=884$ represent the sample mean

$\mu$ population mean (variable of interest)

s=50 represent the sample standard deviation

n=40 represent the sample size

Solution to the problem

Part a

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:

$df=n-1=40-1=39$

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(0.01,39)".And we see that $t_{\alpha/2}=2.426$

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

$884-2.426\frac{50}{\sqrt{40}}=864.82$

$884+2.426\frac{50}{\sqrt{40}}=903.179$

So on this case the 98% confidence interval would be given by (864.82;903.179)

Part b

For this case the upper bound of the confidence interval is lower than 950 so we don't have a significant evidence to conclude that the true mean is 950 per month

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Troyanec [42]

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