Answer:
Step-by-step explanation:
(2b+-7)+5(b+-3)
Multiply 5 throughout the second parentheses
(2b-7)+(5b-15)
Add 2b and 5b
7b-7-15
Subtract -7 and -15
7b-22
Answer:
48
Step-by-step explanation:
So your equation is: 5y3 - 21 + 6y2 + (25 ÷ 5) and y = 2.
Evaluate for y=2
5(23)−21+6(22)+
25
5
5(23)−21+6(22)+
25
5
=48.
So the answer is 48.
For this case, the first thing you should do is define a variable.
We have then:
g: unknown number.
We now write the expression in algebraic form.
the sum of 5 and a number:
g + 5
Answer:
An expression that means the sum of 5 and a number is:
C. g + 5
The answer to your question is 1.70
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- 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 ![s = \sqrt{\frac{p(1 - p)}{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281%20-%20p%29%7D%7Bn%7D%7D)
In this problem:
- Sample of 500 customers, hence
.
- Amazon believes that the proportion is of 70%, hence
![p = 0.7](https://tex.z-dn.net/?f=p%20%3D%200.7)
The <u>mean and the standard error</u> are given by:
![\mu = p = 0.7](https://tex.z-dn.net/?f=%5Cmu%20%3D%20p%20%3D%200.7)
![s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.7(0.3)}{500}} = 0.0205](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281%20-%20p%29%7D%7Bn%7D%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B0.7%280.3%29%7D%7B500%7D%7D%20%3D%200.0205)
The probability is the <u>p-value of Z when X = 0.68</u>, hence:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{0.68 - 0.7}{0.0205}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0.68%20-%200.7%7D%7B0.0205%7D)
![Z = -0.98](https://tex.z-dn.net/?f=Z%20%3D%20-0.98)
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