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2 years ago

Geometry Isosceles and Equilateral Triangles Plsss help me ASAP!!!!

Mathematics

1 answer:

CaHeK987 [17]2 years ago

5 0

Answer:

1. x = 10

(triangle has equal side lengths)

2. x = the same number inside the triangle

(is that a 40 or 80?)

3. x = 34

(is that a 34?)

4. x = the number shown on the other side of the triangle

(is that a 55?)

5. if the given number is 8 then x = 11

(11-3=8)

6. x = 11

(2x-15=7)

7. x = 5

(65+5=70)

8. x = 10

(5x-10=40)

9. m<2 = 9

(7x-3=60)

10. m<2 = 8

(x+72=80)

hope this helps!

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Answer:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=12.8$

The sample deviation calculated $s=4.324 \approx 4.3$

$12.8-4.604\frac{4.324}{\sqrt{5}}=3.896$

$12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704$

So on this case the 99% confidence interval would be given by (3.896;21.704)

Step-by-step explanation:

Data: 10,9,20,13,12

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".

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

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n 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 mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=12.8$

The sample deviation calculated $s=4.324$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first 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 value. The excel command would be: "=-T.INV(0.005,4)".And we see that $t_{\alpha/2}=4.604$

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

$12.8-4.604\frac{4.324}{\sqrt{5}}=3.896$

$12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704$

So on this case the 99% confidence interval would be given by (3.896;21.704)

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