Answer:
C.
x=500$
y=0.2×500+300
y=100+300
y=400
I let 500$=x because they were spent by the customers
hope this helps you
have a great day :)
Answer:
<em>12.99</em>
Step-by-step explanation:
<em>a²+b²=c²</em>
<em>a²+b²=c²√91²+√78²=c²</em>
<em>a²+b²=c²√91²+√78²=c²√8281+√6084=c²</em>
<em>a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²</em>
<em>a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²169=c²</em>
<em>a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²169=c²c=√169</em>
<em>a²+b²=c²√91²+√78²=c²√8281+√6084=c²91+78=c²169=c²c=√169 =13</em>
(8*1) + (5*0.1) + (1*0.01) + (7*0.001).
8.517 in word form is eight and five hundred seventeen thousandths.
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.
Answer:
79
Step-by-step explanation: