If Triangle D K L is congruent to triangle P V X, which of the following statements must also be true? Check all that apply.

Mathematics

2 answers:

Brums [2.3K]3 years ago

5 0

Answer:

A, B And D.

Step-by-step explanation:

I just answered and these were the right choices

sesenic [268]3 years ago

4 0

If the triangle is an equilateral, it should be all of them. it depends on what type it is

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What is the mixed number of 7/4

Nady [450]

7/4 = (4 + 3)/4 = 4/4 + 3/4 = 1 + 3/4 = 1 3/4

$\boxed{\dfrac{7}{3}=1\dfrac{3}{4}}$

5 0

3 years ago

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The proprietor of a boutique in New York wanted to determine the average age of his customers. A random sample of 25 customers r

jolli1 [7]

Answer:

$28-2.064\frac{10}{\sqrt{25}}=23.872$

$28+2.064\frac{10}{\sqrt{25}}=32.128$

So on this case the 95% confidence interval would be given by (23.9;32.1)

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".

$\bar X=28$ represent the sample mean

$\mu$ population mean (variable of interest)

s=10 represent the sample standard deviation

n=25 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:

$df=n-1=25-1=24$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,24)".And we see that $t_{\alpha/2}=2.064$