24
i'm pretty sure of this, if those are the four possible answers.
Answer:
12
Step-by-step explanation:
its not 12 im just trying to make an account
x(1.06)=5.25
The reason it is 1.06 is because it is the original cost plus tax. You can divide each side by 1.06 to get x roughly equals 4.95.
So $4.95 is roughly what the camera costs rounded to the nearest cent.
I think it is 1200 people. Because if 40% are female and there are 480 females, then do 480 x 2 and that equals 960 then half of 480 is 240 so add 960 to 240 and you get 1200 as the total amount of people there. I'm not sure if it's right, but it should be.
Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.
<h3>Normal Probability Distribution</h3>
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, the sampling distribution of sample means of size n has standard deviation
.
In this problem:
- The mean is of 660, hence
.
- The standard deviation is of 90, hence
.
- A sample of 100 is taken, hence
.
The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</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{670 - 660}{9}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B670%20-%20660%7D%7B9%7D)
![Z = 1.11](https://tex.z-dn.net/?f=Z%20%3D%201.11)
has a p-value of 0.8665.
1 - 0.8665 = 0.1335.
0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.
To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213