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:
11 ≤ n
Explanation:
More than indicates the minimum value, so you use the greater than or equal to.
The above answer is written in reverse, which the exact same result.
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Answer:
P = $300
r = 0.15
n = 12
$544.61 (to the nearest cent)
$524.70 (to the nearest cent)
Step-by-step explanation:
P = principal amount = $300
r = annual interest rate in decimal form = 15% = 15/100 = 0.15
n = number of times interest is compounded per unit t = 12
<u>How much she'll owe in 4 years</u>
P = 300
r = 0.15
n = 12
t = 4
= $544.61 (to the nearest cent)
<u>Yearly compounding interest rate</u>
<u>How much she'll owe in 4 years at yearly compounding interest</u>
= $524.70 (to the nearest cent)
