Use the lengths of corresponding sides
so answer is 16:40 which simplifies to 2:5.
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.
We can actually undsitribute the y

divide both sides by


times the right side by (ab/ab)

y=ab
Answer:7.5%
Step-by-step explanation:
decrease the number of visitors from Alec´s website from February to the next month.
$94 would be the correct answer. Hope this helps.