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.
Let's say the smaller number is x. Then, we can represent the larger number as x + 1. The +1 means that it's both larger, because adding makes a number larger, and consecutive, because it's 1 greater or less.
Then, we can write an equation. The sum of the two numbers is x + (x + 1), and it's equal to -165.
x + (x + 1) = -165
2x + 1 = -165
2x = -166
x = -83
Since x = -83, the smaller number is -83.
Step-by-step explanation:
fg(x) =
=
=
Here you go! Hope this helps!
You need to add 2+3+4
divid into 1/3
and your answer is 0.037 longated
