I don't get how to do this, can someone help?

A rectangular patio has a length of 12 1/2 feet and an area of 103 1/8 square feet. What is the width of the patio?

Rina8888 [55]

Length of the rectangular patio = 12 1/2 feet
      = 25/2 feet
Area of the rectangular patio = 103 1/8 square feet
   = 825/8 square feet
Let us assume the width of the rectangular patio = x feet
Then
Area of the rectangular patio = Length * Width
825/8 = (25/2) * x
25x/2 = 825/8
25x = (825 * 2)/8 feet
25x = 825/4 feet
x = 825/(4 * 25) feet
   = 33/4 feet
   = 8 1/4 feet
So the width of the rectangular patio is 8 1/4 feet. I hope the procedure is clear enough for you to understand.

6 0

3 years ago

Identify the slope and y−intercept of the line with the equation 18x − 6y = 12.

DaniilM [7]

And where is the graph? I'll change my answer.

5 0

3 years ago

(8c+8)–(c+3)<br> answer-

otez555 [7]

Answer:

7c + 5

Step-by-step explanation:

(8c+8)–(c+3)

8c - c = 7c

8 - + 3 = 5

7c + 5

6 0

3 years ago

Read 2 more answers

Armando and Bianca are 290 feet apart when they start walking toward one another. Bianca walks twice as fast as

aivan3 [116]

9514 1404 393

Answer:

d = 290 -3x

Step-by-step explanation:

For each x feet that Armondo travels toward Bianca, Bianca travels 2x feet toward Armondo. The net effect is that the distance between them decreases by x+2x = 3x feet. The distance starts at 290 feet when x=0, so the distance can be described by ...

d = 290 -3x

3 0

2 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$