Answer:
The percent of callers are 37.21 who are on hold.
Step-by-step explanation:
Given:
A normally distributed data.
Mean of the data,
= 5.5 mins
Standard deviation,
= 0.4 mins
We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.
Lets find z-score on each raw score.
⇒
...raw score,
=
⇒ Plugging the values.
⇒
⇒
For raw score 5.5 the z score is.
⇒
⇒
Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.
We have to work with P(5.4<z<5.8).
⇒ 
⇒ 
⇒
⇒
and
.<em>..from z -score table.</em>
⇒ 
⇒
To find the percentage we have to multiply with 100.
⇒ 
⇒
%
The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21
Answer:
the answer is A
Step-by-step explanation:
Answer: See explanation
Step-by-step explanation:
From the equation, we are informed that Mr. Robison rides 136 miles per hour on his bike ride and that he wants to ride more than 360 miles at the same rate during his next bike ride.
The inequality that represents all possible values for b, the number of hours Mr. Robison must bike will be:
136b > 360
b > 360/136
b > 2.65
Therefore, Me Robinson must bike for more than 2.65 hours
I believe the answer would be 3/40