Multiply the area of the circle by the percentage:
60 square units x 50% = 60 x 0.50 = 30
50% 0f the circle id 30 square units.
Using the <u>normal distribution and the central limit theorem</u>, it is found that there is a 0.1635 = 16.35% probability of a sample result with 68% or fewer returns prior to the third day.
In a normal distribution with mean
and standard deviation
, the z-score of a measure X is given by:

- It 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, which is the percentile of X.
- By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportions has mean
and standard error 
In this problem:
- Sample of 500 customers, hence
.
- Amazon believes that the proportion is of 70%, hence

The <u>mean and the standard error</u> are given by:


The probability is the <u>p-value of Z when X = 0.68</u>, hence:

By the Central Limit Theorem



has a p-value of 0.1635.
0.1635 = 16.35% probability of a sample result with 68% or fewer returns prior to the third day.
A similar problem is given at brainly.com/question/25735688
2W + 2L = 820
<span>LW = 42,000 or L = 42,000/W </span>
<span>substitute second equation into first equation: </span>
<span>2W + 2(42,000/W) = 820 </span>
<span>2W + 84,000/W = 820 </span>
<span>2W^2/W + 84,000/W = 820W/W </span>
<span>2W^2 - 820W + 84,000 = 0 </span>
<span>quadratic formula: </span>
<span>W = [820 +/- SQR(672,400 - 4(2)(84,000))]/2(2) </span>
<span>W = [820 +/- 20]/4 </span>
<span>W = 200, 210 </span>
<span>using second equation: </span>
<span>L = 42,000/200, 42,000/210 = 210, 200 </span>
<span>The dimensions of the parking lot are 200ft by 210ft.</span>
Area of this rectangle
4×3=12cm^2
formula
long side×short side=area of this rectangle
8 + 6.2 + 4a - 9a
(8 + 6.2) + (4 - 9)a
14.2 - 5a is your answer.