Ask question

Ask question

All categories

2 years ago

10

A garden has width of

le" class="latex-formula"> and length 7 $\sqrt13$ . What is the perimeter of the garden in simplest radical form?
a. 14 $\sqrt13$
b. 16 $\sqrt13\\$
c. 91
d. 8 $\sqrt13$

If you could show work that would be wonderful, but just the answer is greatly apricated as well.

1 answer:

inessss [21]2 years ago

4 0

Answer:

perimeter of the garden=2(length+width)

=2(7√13+√13)

=2(8√13)

=16√13

You might be interested in

Calculate the slope of the line that passes through (3,2) and (−7,4)

34kurt

Answer:

-1/5

Step-by-step explanation:

It has a rise of 2 and a run of -10, therefore it is a slope of -2/10 or -1/5

8 0

3 years ago

Read 2 more answers

1. Use the figure below to answer the questions.<br><br> (a) Solve for x.

Dmitry [639]

Let's see what to do buddy...

_________________________________

The sum of the angles of each n-sided figure is found in the following equation :

$(n - 2) \times 180$

The question's figure is 5-sindes figure

so the sum the angles equal :

$(5 - 2) \times 180 = 3 \times 180 = 540 \\$

So we have :

$90 + 107 + 144 + 138 + x = 54 0 \\$

$479 + x = 540$

Subtract the sides of the equation minus 479

$x = 540 - 479$

$x = 61$

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

5 0

3 years ago

Solve 24x + 2y = 48 for x

drek231 [11]

Isolate x.
24x = 48 - 2y

Divide by 24.
x = 2 - y/12

3 0

3 years ago

Read 2 more answers

(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

...........................

9966 [12]

The answer to your question would be A

7 0

3 years ago