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:Your left hand side evaluates to:
m+(−1)mn+(−1)m+(−1)mnp
and your right hand side evaluates to:
m+(−1)mn+(−1)m+np
After eliminating the common terms:
m+(−1)mn from both sides, we are left with showing:
(−1)m+(−1)mnp=(−1)m+np
If p=0, both sides are clearly equal, so assume p≠0, and we can (by cancellation) simply prove:
(−1)(−1)mn=(−1)n.
It should be clear that if m is even, we have equality (both sides are (−1)n), so we are down to the case where m is odd. In this case:
(−1)(−1)mn=(−1)−n=1(−1)n
Multiplying both sides by (−1)n then yields:
1=(−1)2n=[(−1)n]2 which is always true, no matter what n is
Top Right: 1 3/4 divided by 1/10
Answer: The linear equations intersect at 5 , -4
