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3 years ago

Suppose LMN=PQR Which congruency statements are true

Mathematics

1 answer:

zysi [14]3 years ago

6 0

A, D, E and F are all true statements.

When looking for congruence in these type of problems, you have to look for the order of the letters in the original listing. Since they are listed LMN and PQR, this means we have to focus on the placement of those letters. Since L is the first listed in the sequence, it is congruent with P, which is also listed first. M is congruent with Q since they are both in the middle, and N and R are congruent since they are both at the end.

Now to find true statements, you need to make sure everything matches up in the statement.

Let's take A for instance. It states that MN = QR. Now we know M is in the middle and N is at the end. Since Q is in the middle and R is at the end, they match and are therefore true.

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Maksim231197 [3]

Answer:

The 90% confidence interval would be given by (60.09;69.91)

We are 90% confident that the true mean for the speeds of vehicles traveling on a highway is between 60.09 and 69.91 miles per hour.

Step-by-step explanation:

1) 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".

$\bar X =65$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=9 represent the sample standard deviation

n=11 represent the sample size

2) Confidence interval

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:

$df=n-1=11-1=10$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,10)".And we see that $t_{\alpha/2}=1.81$

Now we have everything in order to replace into formula (1):

$65-1.81\frac{9}{\sqrt{11}}=60.09$

$65+1.81\frac{9}{\sqrt{11}}=69.91$

So on this case the 90% confidence interval would be given by (60.09;69.91)

We are 90% confident that the true mean for the speeds of vehicles traveling on a highway is between 60.09 and 69.91 miles per hour.

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3 years ago