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Delicious77 [7]
3 years ago
10

ohn needs to make a scale drawing of his school building for art class. If the building is 256.25 meters long, and John scales i

t down using a ratio of 25 meters to 1 inch, how long will the building be in the sketch?
Mathematics
1 answer:
kirza4 [7]3 years ago
5 0

the building is 265.25m long and the scale is 1 inch on the drawing is equal to 25 meters of the building. To find the answer lets scale the building. If the real length of the building is 256.25 meters, then the number of inches will be: 256.25/25=10.25 inches.

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d) [-16.03,-3.97]

-16.03 \leq \mu_A -\mu_B \leq -3.97.

Step-by-step explanation:

Notation and 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".  

n_A=18 represent the sample of A

n_B =13 represent the sample of B

\bar x_A =39 represent the mean sample  for A

\bar x_B =49 represent the mean sample for B  

s_A =8 represent the sample deviation for A

s_B =4 represent the sample deviation for B

\alpha=0.01 represent the significance level

Confidence =99% or 0.99

The confidence interval for the difference of means is given by the following formula:  

(\bar X_A -\bar X_B) \pm t_{\alpha/2}\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})} (1)  

The point of estimate for \mu_A -\mu_B is just given by:  

\bar X_A -\bar X_B =39-49=-10  

The appropiate degrees of freedom are df=n_1+ n_2 -2=18+13-2=29

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that t_{\alpha/2}=2.756  

The standard error is given by the following formula:  

SE=\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}  

And replacing we have:  

SE=\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=2.188  

Confidence interval  

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

-10-2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-16.03  

-10+2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-3.97  

So on this case the 99% confidence interval for the differences of means would be given by -16.03 \leq \mu_A -\mu_B \leq -3.97.

d) [-16.03,-3.97]

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