Answer:
The final answer is (2,-1)
(also idek what I did in the photo but I'm positive the answer is (2,-1)
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
You can figure out the answer by doing this:
8% = 2300/x
and do the same for the second one
(and solve)
Answer:
5 1/10 meters
Step-by-step explanation:
18 4/5 can be changed to 18 8/10 so
18 8/10 - 13 7/10 = 5 1/10 meters
The answer is d, hope this helps