Answer:
5.48% of the people in line waited for more than 28 minutes
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean waiting time of 20 minutes with a standard deviation of 5 minutes.
This means that 
What percentage of the people in line waited for more than 28 minutes?
The proportion is 1 subtracted by the p-value of Z when X = 28. So



has a p-value of 0.9452.
1 - 0.9452 = 0.0548.
As a percentage:
0.0548*100% = 5.48%
5.48% of the people in line waited for more than 28 minutes
So begin by doing 360-284=76
90+76=166
180-166=14
X=14
Rule of 90 clockwise rotation
(x,y) --->(y, -x)
C(2,7) ,90 clockwise rotation will be C'(7, -2)
Answer: second option (7, -2)
Hi!
In this problem, you would multiply the constants and then add the exponents. The answer would be (5*5)n^(4+4), or 25n^8.
The answer is A: 25n^8
Answer:
![\displaystyle y = 5x - 92\:or\:y + 12 = 5[x - 16]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%3D%205x%20-%2092%5C%3Aor%5C%3Ay%20%2B%2012%20%3D%205%5Bx%20-%2016%5D)
Step-by-step explanation:
−12 = 5[16] + b
80

To write this in Point-Slope Formula now,
you simply plug the coordinate into the formula with their CORRECT SIGNS because all negative symbols give the OPPOSITE terms of what they really are:
![\displaystyle y + 12 = 5[x - 16]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%20%2B%2012%20%3D%205%5Bx%20-%2016%5D)
* Parallel lines have SIMILAR <em>RATE OF CHANGES</em> [<em>SLOPES</em>], so 5 remains the way it is.
I am joyous to assist you anytime.