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3 years ago

Write an equation of the line passing through the points (3,8) and (-2,-22)

Mathematics

1 answer:

uranmaximum [27]3 years ago

8 0

Answer:

y=6x+8 or y=6x-22

Step-by-step explanation:

use the equation y=mx+b

m is the slope and b is the y-intercept

to find the slope use the equation m = y2-y1 over x2-x1

-22 - 8 / -2 - 3 = -30 / -5 = 6

6 is your slope and for the y-intercept you can choose whichever y-intercept to put into the equation.

HOPE THIS HELPS!!

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(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago

15.09 ÷ 0.01<br><br><br> Plzz help

Vika [28.1K]

Answer:

1509

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

to go to the dog beach, it costs $3 plus $2 per dog. write an equation that shows this situation, where "x" represents the numbe

bixtya [17]

The equation representing this situation would be 3 + 2x = c, where c is the total cost.

To find the cost to bring 12 dogs, you will replace the x with the number 12.

3 + 2 x 12 = 27

The cost would be $27.

8 0

3 years ago

A student made an error when solving this problem. A number divided by 4 is 3 more than –11. What is the number?

fiasKO [112]

C is the right anskwe

4 0

3 years ago

Use elimination to solve the system.

julia-pushkina [17]

Answer:

the answer to your question is (9,3)

4 0

2 years ago