Answer:
B
Step-by-step explanation:
Initial mean = 156, Changed mean = 163.92
So, mean increases. Only condition that satisfies this is B.
Answer:
don't know sorry
Step-by-step explanation:
dont know sorry
Correct Ans:Option A. 0.0100
Solution:We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.
First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:
So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.
Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.
Answer:
100
Step-by-step explanation:
Simplify the following:
4 (12 - 2)^2/4
Hint: | Express 4 (12 - 2)^2/4 as a single fraction.
4 (12 - 2)^2/4 = (4 (12 - 2)^2)/4:
(4 (12 - 2)^2)/4
Hint: | Cancel common terms in the numerator and denominator of (4 (12 - 2)^2)/4.
(4 (12 - 2)^2)/4 = 4/4×(12 - 2)^2 = (12 - 2)^2:
(12 - 2)^2
Hint: | Subtract 2 from 12.
| 1 | 2
- | | 2
| 1 | 0:
10^2
Hint: | Evaluate 10^2.
| 1 | 0
× | 1 | 0
| 0 | 0
1 | 0 | 0
1 | 0 | 0:
Answer: 100
If , then and because lies in the fourth quadrant.
By DeMoivre's theorem,
so the answer is B.