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)
Answer:
(-2,-1)
Step-by-step explanation:
Midpoints are found when you do ([x(subscript A) + x(subscript B)]/2, [y(subscript A) + y(subscript B)]. Therefore your equation (for x of B) should look like this:
-4 = ([-6+x]/2)
After you solve for x, you should have gotten -2 for x.
Then you do the same for y:
2 = ([5+y]/2)
You should get -1. Your final answer should be (-2, -1)
Answer:
For the first row, write 5(0)-1 = 0-1. y is equal to -1.
For the second row, write 5(1)-1 = 5-1. y is equal to 4.
For the third row, write 5(2)-1 = 10-1. y is equal to 9.
For the fourth row, w rite 5(3)-1 = 15-1. y is equal to 14.
Step-by-step explanation:
Hope this helps!
Step-by-step explanation:
We need to find an expression for 
.
We can solve it as follows.
We know that,
So,
or
Hence, this is the required solution.
Answer:
9( 1 + 4w)
Step-by-step explanation:
So if you're simplifying then you need to find the greatest common factor. The GCF is 9.
