Write the equation of the line that passes through

In algebra, we use a(n) __to represent a variable. Answer here

Vlada [557]

Answer:

x or another letter

Step-by-step explanation:

3 0

2 years ago

What is the scale factor from AABC to AUVW?

Hitman42 [59]

A, 1/5 yes hehehehehehe

5 0

2 years ago

2. A cube has a volume of 7 to the third cubic centimeters. What is its surface area? الم

Marizza181 [45]

Answer:

294

Step-by-step explanation:

since the area is 7^3, the cube's side length is 7

SA = 6*7^2 = 294

7 0

2 years ago

What is the equation of the line with an x-intercept of -1 and a y-intercept of 2?

svlad2 [7]

Answer:

y = 2x + 2

Step-by-step explanation:

(a) Slope

The point-slope formula for a straight line is

y₂ - y₁ = m(x₂ - x₁)

The line goes through the points (-1, 0) and (0, 2).

2 – 0 = m[0 – (-1)]

2 = m

(b) y-intercept

y = mx + b

Use the point (0, 2)

2 = 2× 0+ b

2 = 0 + b

b = 2

∴ y = 2x + 2

3 0

3 years ago

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:

a)

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

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

b)

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.

c)

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

3 0

2 years ago