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katen-ka-za [31]
3 years ago
5

SOMEONE PLZZ HELP MEEEE BRAINLIST.

Mathematics
1 answer:
adell [148]3 years ago
3 0

9514 1404 393

Answer:

  A, C, D

Step-by-step explanation:

All three triangles can be proved congruent, so corresponding angles are congruent.

  ∆JKL ≅ ∆MNO ≅ ∆PQR

__

So, the statements that are true are ...

  A) rigid motion can map ∆JKL to ∆MNO

  C) ∆JKL can be proved congruent to ∆PQR

  D) ∠N can be proved congruent to ∠Q

_____

<em>About the other choices</em>

B. When triangles are congruent, there is some rigid motion that maps one to the other. It is false to say there is none such.

E. ∠L has a measure of 65° (not 55°), as do the corresponding angles in the other triangles.

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The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confide
blsea [12.9K]

Answer:

38.898 \leq \sigma^2 \leq 2792.356

Now we just take square root on both sides of the interval and we got:

6.237 \leq \sigma \leq 52.843

Step-by-step explanation:

Data given and notation

34,59,61,71,59

We can calculate the sample standard deviation with the following formula:

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

s= \sqrt{s^2}

s=12.021 represent the sample standard deviation

\bar x represent the sample mean

n=5 the sample size

Confidence=99% or 0.99

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 mean or variance 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.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=5-1=4

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 values.  

The excel commands would be: "=CHISQ.INV(0.005,4)" "=CHISQ.INV(1-0.005,4)". so for this case the critical values are:

\chi^2_{\alpha/2}=14.860

\chi^2_{1- \alpha/2}=0.207

And replacing into the formula for the interval we got:

\frac{(4)(12.021)^2}{14.860} \leq \sigma^2 \leq \frac{(4)(12.021)^2}{0.207}

38.898 \leq \sigma^2 \leq 2792.356

Now we just take square root on both sides of the interval and we got:

6.237 \leq \sigma \leq 52.843

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