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3 years ago

I NEED AN ANSWER IMMEDIATELY!

1 answer:

3 0

Answer:
area = w(w+5) = w^2 + 5w square units

Explanation:
The pool has a rectangular shape. Area of rectangle can be calculated by multiplying its length and width.
We are given that:
length = w + 5
width = w
Therefore:
Area of pool = length * width
= (w+5) * w
= w(w+5) = w^2 + 5w square units

Hope this helps :)

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HELPPP PLSSSS ITS ALMOST DUE IN 1 HOUR WILL GIVE BRAINLIEST THANKS AND 5 STARSSS

svp [43]

Answer:Its been 5 hours. so 85 - 82 = X. X times 0.6 = 5 then translate to 5 hours.

Step-by-step explanation:

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3 years ago

How do I solve this problem?

lara31 [8.8K]

Answer:

x = 12

Step-by-step explanation:

(whole secant) x (external part) = (whole secant) x (external part)

(6+x) *6 = (4+23) * 4

(6+x) *6 = 27*4

6(6+x) = 108

Divide each side by 6

6+x =18

Subtract 6 from each side

6+x-6 =18-6

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4 years ago

Read 2 more answers

If 4 pups are born, what is the probability that all are male

Neporo4naja [7]

Answer:

half

Step-by-step explanation:

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3 years ago

In 1998, as an advertising campaign, the Nabisco Company announced a "1000 Chips Challenge," claiming that every 18-ounce bag of

Phoenix [80]

Answer:

A 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

Step-by-step explanation:

We are given that statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies and counted the chocolate chips.

Some of their data are given below; 1219, 1214, 1087, 1200, 1419, 1121, 1325, 1345, 1244, 1258, 1356, 1132, 1191, 1270, 1295, 1135.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean number of chocolate chips = $\frac{\sum X}{n}$ = 1238.2

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 94.3

n = sample of car drivers = 16

$\mu$ = population mean number of chips in a bag

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of