The first thing you should do is use the method of variable separation. Then, integrate both sides to find the solution.
In this case the answer is
<span>D. - 1/y = 3x4/3 + C</span>
I attach the solution
Answer:
Step-by-step explanation:
Since the length of time taken on the SAT for a group of students is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - u)/s
Where
x = length of time
u = mean time
s = standard deviation
From the information given,
u = 2.5 hours
s = 0.25 hours
We want to find the probability that the sample mean is between two hours and three hours.. It is expressed as
P(2 lesser than or equal to x lesser than or equal to 3)
For x = 2,
z = (2 - 2.5)/0.25 = - 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.02275
For x = 3,
z = (3 - 2.5)/0.25 = 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.97725
P(2 lesser than or equal to x lesser than or equal to 3)
= 0.97725 - 0.02275 = 0.9545
