Answer:
- <em>m</em> =
- <em>μ</em> = 20
- <em>σ </em>= 20
The probability that a person is willing to commute more than 25 miles is 0.2865.
Step-by-step explanation:
Exponential probability distribution is used to define the probability distribution of the amount of time until some specific event takes place.
A random variable <em>X</em> follows an exponential distribution with parameter <em>m</em>.
The decay parameter is, <em>m</em>.
The probability distribution function of an Exponential distribution is:

<u>Given</u>: The decay parameter is, 
<em>X</em> is defined as the distance people are willing to commute in miles.
- The decay parameter is <em>m</em> =
. - The mean of the distribution is:
. - The standard deviation is:
Compute the probability that a person is willing to commute more than 25 miles as follows:

Thus, the probability that a person is willing to commute more than 25 miles is 0.2865.
Answer:
-6
Step-by-step explanation:
Given that :
we are to evaluate the Riemann sum for
from 2 ≤ x ≤ 14
where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.
The Riemann sum can be computed as follows:

where:

a = 2
b =14
n = 6
∴



Hence;

Here, we are using left end-points, then:

Replacing it into Riemann equation;






Estimating the integrals, we have :

= 6n - n(n+1)
replacing thevalue of n = 6 (i.e the sub interval number), we have:
= 6(6) - 6(6+1)
= 36 - 36 -6
= -6
Answer:
8 inch is your best buy
Step-by-step explanation:
8inch diameter =10 dollars
5 inch radius = 12 dollars
so if we get each equal to the diameter it should be 10 inch and that would be 24 dollars vs . 8 in for 10 dollars. So best buy would be 8 inch
It’s would be 160 km since 160/80x60=120 min and 160/120x60=80min
Answer:
39
Work:
(x+5)/2=22
We will reverse this equation.
We will start by moving the 2 through inverse operations.
22x2=44
Now we have the equation
x+5=44
We will subtract 5
44-5=39
x=39
the unknown variable equals 39
Kali had 39 CDs on Monday