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.
Answer:
I guess it's -2
Step-by-step explanation:
6-8x = 5x - 10x + 12
∴ 10x - 8x - 5x = 12 - 6
∴ 10x- 13x = 6
∴ - 3x = 6
∴ x = -6\3
∴ x = -2
Answer:
x=11
Step-by-step explanation:
7x-9-4(x-1)=28
7x-9-4x+4=28
3x-5=28
3x=33
x=11
