PLZ HELP!!!!!!!!!!!!!!

Mathematics

1 answer:

Sladkaya [172]3 years ago

3 0

2(3.14)(.75)(3)+2(3.14)(.75^2)
14.13+3.5325
17.8 approx
Answer would be A . I just used 3.14 instead of pi symbol on calc so answer is slightly off

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uncle percy sold t shirts at the show. now he only has 10% of the t shirts he started with. if he has 25 t shirts now , how many

kramer

He had 250 shirts to begin with.

Let x be the number of shirts he began with. He has 10%, or 0.10 of the shirts now he had then. This equals 25:

0.1x = 25

Divide both sides by 0.1:
0.1x/0.1 = 25/0.1
x = 250

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Six measurements were made of the magnesium ion concentration (in parts per million, or ppm) in a city's municipal water supply,

myrzilka [38]

Answer:

$163.83-4.03\frac{20.094}{\sqrt{6}}=130.77$

$163.83+4.03\frac{20.094}{\sqrt{6}}=196.89$

So on this case the 99% confidence interval would be given by (130.77;196.89)

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".

Data: 175 177 175 180 138 138

We can calculate the mean and the deviation from these data with the following formulas:

$\bar X= \frac{\sum_{i=1}^n x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

$\bar X=163.83$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=20.093 represent the sample standard deviation

n=6 represent the sample size

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:

$df=n-1=6-1=5$

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(0.005,5)".And we see that $t_{\alpha/2}=4.03$