Answer:
Mailing preparation takes 38.29 min max time to prepare the mails.
Step-by-step explanation:
Given:
Mean:35 min
standard deviation:2 min
and 95% confidence interval.
To Find:
In normal distribution mailing preparation time taken less than.
i.eP(t<x)=?
Solution:
Here t -time and x -required time
mean time 35 min
5 % will not have true mean value . with 95 % confidence.
Question is asked as ,preparation takes less than time means what is max time that preparation will take to prepare mails.
No mail take more time than that time .
by Z-score or by confidence interval is
Z=(X-mean)/standard deviation.
Z=1.96 at 95 % confidence interval.
1.96=(X-35)/2
3.92=(x-35)
X=38.29 min
or
Confidence interval =35±Z*standard deviation
=35±1.96*2
=35±3.92
=38.29 or 31.71 min
But we require the max time i.e 38.29 min
And by observation we can also conclude the max time from options as 38.29 min.
I suppose the integral is
The integration region corresponds to a sector of a cirlce with radius 5 subtended by a central angle of π/4 rad. We can capture this region in polar coordinates by the set
Then 
, 
, and 
. So the integral becomes
Answer:
32
Step-by-step explanation:
77 quarters and dimes
q + d = 77
subtract q from both sides
d = 77 - q
---------------------------
A collection of quarters and dimes is worth $12.50 (1250 cents)
25q + 10d = 1250
substitute for d
25q + 10(77 - q) = 1250
distribute
25q + 770 - 10q = 1250
Combine like terms
15q + 770 = 1250
subtract 770 from both sides
15q = 480
divide both sides by 15
q = 32
Answer:
x=6
Step-by-step explanation:
First, we combine like terms.
6x=2x+24
6x-2x=24
4x=24
Then we divide both sides by 4.
x=6
