Based on your plot points, the best answer seems to be B) $5.00
Step-by-step explanation:
m< AOC = 90°
m< AOB +m<BOC = 90°
6x-12+3x +30 = 90°
9x +18° = 90°
9x = 90- 18° = 72°
x = 8°
m<AOB =6(8)-12 = 48-12 = 36°
Answer:
A customer who sends 78 messages per day would be at 99.38th percentile.
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.
Average of 48 texts per day with a standard deviation of 12.
This means that 
a. A customer who sends 78 messages per day would correspond to what percentile?
The percentile is the p-value of Z when X = 78. So



has a p-value of 0.9938.
0.9938*100% = 99.38%.
A customer who sends 78 messages per day would be at 99.38th percentile.
This is the answer. 4.375x
Answer:
9x/4= -9
Mutiply by 4 for both sides.
4(9x/4) cross out 4 and 4, divide by 4 and then becomes 9x. (-9)(4)= -36
9x= -36
Divide by 9 for both sides
9x/9= -36/9
x= -4