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Temka [501]
3 years ago
11

On a coordinate plane, point D is 1 unit up. Point C is 2 units to the right and 10 units up. Point F is 8 units to the right an

d 6 units up. Point G is 9 units to the right and 2 units up. Point E is 12 units to the right and 12 units up.
Look at the plotted points. Which statements are true? Select all that apply.

Point C is located at (1, 5).

Point D is located at (0, 1).

Point E is located at (12, 6).

Point F is located at (8, 6).

Point G is located at (9, 2).
Mathematics
2 answers:
kotykmax [81]3 years ago
8 0

Answer:

point D is 1 unit up; true

Point C is 2 units to the right and 10 units up; false

Point F is 8 units to the right and 6 units up; true

Point G is 9 units to the right and 2 units up; true

Point E is 12 units to the right and 12 units up; false

Step-by-step explanation:

Nutka1998 [239]3 years ago
7 0

Answer:

The other person is wrong,

Step-by-step explanation:

b, d and e are the correct answers

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a) \frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

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

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b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

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}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}&#10;And in order to find the sample mean we just need to use this formula:&#10;[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample deviation for this case is s=30.23

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

df=n-1=9-1=8

The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

\chi^2_{1- \alpha/2}=1.65

And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

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

19.08 \leq \sigma \leq 66.56

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

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4 years ago
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