Negative 3................
Answer:
E
Step-by-step explanation:
Label the two equations:
4p +3q +2r= 3 -----(1)
-4p -2q -2r= 4 -----(2)
(1) +(2):
4p +3q +2r -4p -2q -2r= 3 +4
Grouping like terms together:
4p -4p +3q -2q +2r -2r= 7
Simplify:
q= 7
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:
- 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:
By the Central Limit Theorem
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
1.2 ?
At f(-6) it looks like it’s 1.2 “(-6, 1.2)
Answer:
2 * (i^6) * (3 * i) =
- 6 i
Step-by-step explanation: