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3 years ago

I’m confused on how to do this

Mathematics

1 answer:

vladimir2022 [97]3 years ago

3 0

Answer:

(6,6) only goes with Line 2

(3,4) goes with neither

(7,2) goes with both

Step-by-step explanation:

Ok to decide if a point is on a line you plug it in. If you get the same thing on both sides, then that point is on that line. If you don't get the same thing on both sides, then that point is not on that line.

Test (6,6) for -5x+6y=-23.

(x,y)=(6,6) gives us

-5x+6y=-23

-5(6)+6(6)=-23

-30+36=-23

6=-23

So (6,6) is not on -5x+6y=-23.

Test (6,6) for y=-4x+30

(x,y)=(6,6) give us

y=-4x+30

6=-4(6)+30

6=-24+30

6=6

So (6,6) is on y=-4x+30.

Test (3,4) for -5x+6y=-23.

(x,y)=(3,4) gives us

-5x+6y=-23

-5(3)+6(4)=-23

-15+24=-23

9=-23

So (3,4) is not on -5x+6y=-23.

Test (3,4) for y=-4x+30.

(x,y)=(3,4) gives us

y=-4x+30

4=-4(3)+30

4=-12+30

4=18

So (3,4) is not on y=-4x+30.

Test (7,2) for -5x+6y=-23.

(x,y)=(7,2) gives us

-5x+6y=-23

-5(7)+6(2)=-23

-35+12=-23

-23=-23

So (7,2) is on -5x+6u=-23.

Test (7,2) for y=-4x+30.

(x,y)=(7,2) gives us

y=-4x+30

2=-4(7)+30

2=-28+30

2=2

So (7,2) is on y=-4x+30

(x,y) Line 1 Line 2 Both Neither

(6,6) *

(3,4) *

(7,2) *

(6,6) only goes with Line 2

(3,4) goes with neither

(7,2) goes with both

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An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

Testing the hypothesis, it is found that:

The null hypothesis is: $H_0: \mu \leq 10$

The alternative hypothesis is: $H_1: \mu > 10$

The critical value is: $t_c = 1.74$

The decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of at most 10 pounds, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of more than 10 pounds, that is:

$H_1: \mu > 10$

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a significance level of 0.05 and 18 - 1 = 17 df.

Hence, using a calculator for the t-distribution, the critical value is: $t_c = 1.74$ .

Hence, the decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Item c:

We have the standard deviation for the sample, hence the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

For this problem, we have that:

$\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18$

Thus, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}$

$t = 1.41$

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

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Andre45 [30]

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Tasya [4]

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scoundrel [369]

Answer:no

Step-by-step explanation: