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2 years ago

11

The length of the hypotenuse of a right triangle is 16 inches. If the length of one leg is 5 inches, what is the approximate len

gth of the other leg? 10.5 inches 11.0 inches 15.2 inches 16.8 inches

1 answer:

Musya8 [376]2 years ago

7 0

Answer:

15.2 inches

Step-by-step explanation:

a^2 + b ^2= c^2

5^2 + b ^2=16^2

b ^2=256 - 25

√b ^2=√231

b= 15.2 inches

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What is the square root of 4

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Square root of a number is 2 numbers that multiply to get 4 that are the same
first find the factors

4=2 times 2
then find the doubles and write only one from each pair
√4=2

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36=2 times 2 times 3 times 3
the doubles are 2 and 3 so
√36=2 times 3=6

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The circumference of a circle is 18.84 meters. what is the circles area

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3 years ago

For every 8 red pens on the teachers desk there are 12 pencils. If she has 60 writing implements, does she have enough red pens

gavmur [86]

Answer:No she does not

Step-by-step explanation:

8pens + 12pecils= 20writing inplements

20+8+12=40

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2 years ago

Assume that population mean is to be estimated from the sample described. Use the sample results to approximate the margin of er

Vikentia [17]

Answer:

$ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1$

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".

$\bar X =69$ represent the sample mean for the sample

$\mu$ population mean

s=4.0 represent the sample standard deviation

n=64 represent the sample size

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma)$

The sample mean $\bar X$ have the following distribution:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

The margin of error for the sample mean is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

Since we have the sample standard deviation we can estimate the margin of error like this:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$

We can find the degrees of freedom like this:

$df=n-1=64-1=63$

And we can find the critical value with the following code in excel for example: "=T.INV(0.025,63)" and we got:

$t_{\alpha/2}=\pm 1.998$

And the margin of error would be :

$ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1$

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3 years ago

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kondor19780726 [428]

Answer:

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