The perimeter of a rhombus is 12x + 28 feet. Factor this expression. Then find the length of one side if x=8.

ehidna [41]

To find slope subtract the two points in the coordinate (y). We have 8 and 4 so subtract those two. You get 8-4=4. Now start from the same coordinate expect this time subtract (x). Which is 9-3=6. You have 4/6 simplify and you get 2/3. Your slope is 2/3. Hope this helps

5 0

3 years ago

Caren is making rice and beans. She can spend no more than $10 on the ingredients. Which situation is represented by the inequal

sergij07 [2.7K]

If you are given $10 every 3.5 hours and you want to find the unit rate per hour, you want to solve for how much money you make every hour. You do this by dividing the $10 by the 3.5 hours, which will result in about $2.86 per hour. (This is a rounded value)

5 0

3 years ago

(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

3 years ago

How much do I pay for the loan

Natali5045456 [20]

500 / 100 = 5 5 x 15 = 75 75 x 4 = 300 300 + 500 = 800
After the loan, you will pay $800. Hope this helps :)

4 0

3 years ago

Look at the graph below. what is the slope of the line? a) -1/2 b) 1/2 c) -2 d) 2

il63 [147K]

I believe it is b)
Because you always do rise over the run and if it goes right then it is positive and if it goes left it is negative.