Answer:
0.1587
Step-by-step explanation:
Let X be the commuting time for the student. We know that 
. Then, the normal probability density function for the random variable X is given by
![f(x) = \frac{1}{\sqrt{2\pi(5)^{2}}}\exp[-\frac{(x-\mu)^{2}}{2(5)^{2}}]](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B2%5Cpi%285%29%5E%7B2%7D%7D%7D%5Cexp%5B-%5Cfrac%7B%28x-%5Cmu%29%5E%7B2%7D%7D%7B2%285%29%5E%7B2%7D%7D%5D)
. We are seeking the probability P(X>35) because the student leaves home at 8:25 A.M., we want to know the probability that the student will arrive at the college campus later than 9 A.M. and between 8:25 A.M. and 9 A.M. there are 35 minutes of difference. So,
= 0.1587
To find this probability you can use either a table from a book or a programming language. We have used the R statistical programming language an the instruction pnorm(35, mean = 30, sd = 5, lower.tail = F)
At the end of the year, Juan has 52.71 more than 4 times his balance at the beginning. Okay, let's set this up.
4x + 52.71
(4 times) (52.71 more)
His ending was 172.90, so
4x + 52.71=172.90
4x= 120.19
x= 30.05
He had $30.05 at the beginning of the year.
Hello!
Answer:
First, write 0.63 as 0.63/1
Then, multiply both numerator and denominator by 10 for every number after the decimal point.
0.63x100/1x100=63/100
so your answer is 63/100 :)
Hope this helps
Percent increase=increase/original times 100
increase=26-21=5
original=21
so
5/21 times 100=0.23809523809523809523809523809524 times 100=23.809523809523809523809523809524
round
24%
so 24% increase
