Answer:
95% of the customers have to wait between 10 minutes and 26 minutes
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 18 minutes
Standard Deviation, σ = 4 minute
We are given that the distribution of amount of time is a bell shaped distribution that is a normal distribution.
Empirical Formula:
- Almost all the data lies within three standard deviation from the mean for a normally distributed data.
- About 68% of data lies within one standard deviation from the mean.
- About 95% of data lies within two standard deviations of the mean.
- About 99.7% of data lies within three standard deviation of the mean.
Now, we can write:
Thus, by empirical formula, 95% of the data lies within two standard deviations of the mean.
Thus, 95% of the customers have to wait between 10 minutes and 26 minutes
Answer is a because a^-2 will be changed to 1/a^2
Hope that helps
Answer:
T=50.99
Step-by-step explanation:
Given:
I=3,875.78
P=1,900
R=4% I.e 4/100 , 0.04
T=?
I=PRT
3,875.79=($1,900)(0.04)t
3,875.79=76t
3,875.79/76= t
T=50.99
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).
