Answer: Last Option

Step-by-step explanation:
Cramer's rule says that given a system of equations of two variables x and y then:


For this problem we know that:

Solving we have:


Solving we have:


Solving we have:

Finally


Answer:
The answer is 11, hope i helped!
Answer:
0.3333 = 33.33% probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.
Step-by-step explanation:
A distribution is called uniform if each outcome has the same probability of happening.
The uniform distributon has two bounds, a and b, and the probability of finding a value between c and d is given by:

A particular employee arrives at work sometime between 8:00 a.m. and 8:30 a.m.
We can consider 8 am = 0, and 8:30 am = 30, so 
Find the probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.
Between 15 and 25, so:

0.3333 = 33.33% probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.
Let p and m represent the numbers of miles in the plains and mountains, respectively.
.. p + m = 300 . . . . . . . the whole trip was 300 miles
time = distance/speed, so the trip time can be written as
.. p/90 + m/37.5 = 3 48/60
.. 5p + 12m = 1710 . . . . . . . . . . multiply by 450 to put in standard form
You can solve these two equations by any of several means. The variable p can be eliminated by subtracting 5 times the first from the second.
.. (5p +12m) -5(p +m) = 1710 -5*300
.. 7m = 210
.. m = 30
30 miles of the trip was through the mountains.