When Colton commutes to work, the amount of time it takes him to arrive is normally distributed with a mean of 41 minutes and a standard deviation of 3 minutes. What percentage of his commutes will be between 33 and 35 minutes, to the nearest tenth?

And we can find the probability of interest with this difference

And if we use the normal standard table or excel we got:

And if we convert the probability to a % we got 1.896% and rounded to the nearest tenth we got 1.9 %

Step-by-step explanation:

Let X the random variable that represent the times to conmutes to work of a population, and for this case we know the distribution for X is given by:

Where and

We are interested on this probability

And we can solve the problem using the z score formula given by:

And using this formula we got:

And we can find the probability of interest with this difference

And if we use the normal standard table or excel we got:

And if we convert the probability to a % we got 1.896% and rounded to the nearest tenth we got 1.9 %

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

x= 3^y

step-by-step explanation: