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)
x+7
------------------- =
x^2+4x -21
x+7
= -------------------
(x+7)(x-3)
1
= ---------- where x is not equal -7 and 3
x - 3
answer is A. first choice
Step-by-step explanation:
the correct answer is in there
Answer:
y = 2/5 x - 3/5
Step-by-step explanation:
2x = 5y + 3
Rewrite
5y + 3 = 2x
Subtract 3 from both sides
5y = 2x - 3
Divide both sides by 5
y = 2/5 x - 3/5
