All categories

2 years ago

The perimeter of the aqua rectangle is 72 feet. What is the measure of the rectangle’s longest side?

Mathematics

1 answer:

dimulka [17.4K]2 years ago

8 0

Answer:

Step-by-step explanation:

72 = x + 3x - 4

72 = 4x - 4

72 + 4 = 4x - 4 + 4

76 = 4x

76/4 = 4x/4

x = 19

3(19) - 4 = 53

Hope this helps!

You might be interested in

Gabriel invested $900 in an account paying an interest rate of 3 1/4 % compounded daily. Daniel invested $900 in an account payi

Helen [10]

Answer:

$40

Step-by-step explanation:

The earnings of Gabriel can be calculated with the formula:

P = Po * (1 + r/n)^(nt)

Where P = 900, r = 0.0325, n = 365 and t = 19

So we have:

P = 900 * (1 + 0.0325/365)^(365*19) = $1,668.81

The earnings of Daniel can be calculated with the formula:

P = Po * e^(rt)

Where Po = 900, r = 0.03375 and t = 19

So we have:

P = 900 * e^(0.03375*19) = $1,708.97

So the amount that Daniel will have more than Gabriel is:

1708.97 - 1668.81 = $40.16

Rounding to nearest dollar, the difference is $40

3 0

2 years ago

Nakita bought items at a grocery store.  She bought 2 boxes of crackers for $3.50 each.  She used a coupon for $0.80 off the p

olga nikolaevna [1]

Answer: $12.65

Step-by-step explanation: sikeeeeeeeeeeee

4 0

2 years ago

A number c minus 4.3 is no less than 16. write this word sentence as an inequality.

aksik [14]

Answer:

c - 4.3 > 16

(hope this helps)

6 0

2 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

2 years ago

May I please get help with this question?

Julli [10]

The answer should be 42 times.

5 0

3 years ago

Read 2 more answers