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3 years ago

What is the length of line segment KJ?

Mathematics

1 answer:

Karolina [17]3 years ago

5 0

Please consider the attached file.

We can see that triangle JKM is a right triangle, with right angle at M. Segment KM is 6 units and segment MJ is 3 units. We can also see that KJ is hypotenuse of right triangle.

We will use Pythagoras theorem to solve for KJ as:

$KJ^2=KM^2+MJ^2$

$KJ^2=6^2+3^2$

$KJ^2=36+9$

$KJ^2=45$

Now we will take positive square root on both sides:

$\sqrt{KJ^2}=\sqrt{45}$

$KJ=\sqrt{9\cdot 5}$

$KJ=3\sqrt{5}$

Therefore, the length of line segment KJ is $3\sqrt{5}$ and option D is the correct choice.

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|×|=×<br>is the absolute value of x always, sometimes, or never x

il63 [147K]

Answer:

Sometimes

Step-by-step explanation:

I’ll give you an example so you can understand:

Let’s say x is 4. So plug 4 into the problem:

|4|=4 → This is a very true statement, where the absolute value of 4 is equal to 4.

Now, let’s say x is -7. So plug -7 into the problem:

|-7|=-7 → This is a false statement because it’s saying that the absolute value of -7 is -7 which is very untrue.

So |x|=x only works for positive numbers, but not negative numbers. Therefore, |x|=x is the absolute value of x <u>sometimes.</u>

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Hope this helps and answers your question! :)

6 0

3 years ago

A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec

VMariaS [17]

Answer:

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

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".

$\bar X=19.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 represent the sample size

Solution to the problem

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:

$df=n-1=9-1=8$

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(0.005,8)".And we see that $t_{\alpha/2}=$