All categories

3 years ago

Find the slope of the line that passes through: (-4,7) and (-6, -4)

Mathematics

1 answer:

Yuri [45]3 years ago

6 0

Answer:

$\frac{11}{2}$ or 5.5

Step-by-step explanation:

In order to find the slope of a line using two points on a graph, you must use

$y_2-y_1/x_2-x_1$

Pluging in the points (-4,7) and (-6, -4) to the formula you will get:

$\frac{-4-7}{-6+4} =\frac{-11}{-2} =5.5$

The equation for the line would be $\frac{11}{2}$ x+29

You might be interested in

Find unit rate for 51 in 14 min.

wariber [46]

Its going to be .27ths of an inch for 1 minute.

7 0

4 years ago

Read 2 more answers

Sarah starts a jewelry making business. she borrows the money for the supplies to start. she borrows $1,500.00 for 2 years at a

kolezko [41]

Answer:

$260

Step-by-step explanation:

($1500×.12×2)=$360

4 0

1 year ago

UCF believes that the average time someone spends in the gym is 56 minutes. The university statistician takes a random sample of

11111nata11111 [884]

Answer:

We conclude that the average time someone spends in the gym is different from 56 minutes.

Step-by-step explanation:

We are given that UCF believes that the average time someone spends in the gym is 56 minutes.

The university statistician takes a random sample of 32 gym goers and finds the average time of the sample was 50 minutes. Assume it is known the standard deviation of time all people spend in the gym is 8 minutes.

Let $\mu$ = population average time someone spends in the gym

SO, Null Hypothesis, $H_0$ : $\mu$ = 56 minutes {means that the average time someone spends in the gym is 56 minutes}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 56 minutes {means that the average time someone spends in the gym is different from 56 minutes}

The test statistics that will be used here is One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average time someone takes in the gym = 50 min

$\sigma$ = population standard deviation = 8 minutes

n = sample of gym goers = 32

So, test statistics = $\frac{50-56}{\frac{8}{\sqrt{32} } }$

= -4.243

Since in the question we are not given the level of significance so we assume it to b 5%. Now at 5% significance level, the z table gives critical value between -1.96 and 1.96 for two-tailed test. Since our test statistics does not lie within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the average time someone spends in the gym is different from 56 minutes.

4 0

3 years ago

PLEASE HELP QUICKKKKKKKKKKK

ddd [48]

Answer:

$\frac{0.04}{n}$ , or $\frac{1}{25n}$