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:
x = -2/9
Step-by-step explanation:
8x+1=-x-1
8x = -x -2
9x = -2
x = -2/9
Answer:
H
Step-by-step explanation:
The amount of hours he has worked would be x 8$
So let's break this question down.
You need a bit of both, right? Well we need to calculate how many multiple choices we can do whilst having true/false questions, and not going over or under on points.
So, let's pick a random number, 5. So let's do 8x5, that equals 40.
ohmygoditsrightiwasseriouslyjustguessing
Ok so now we have our number, 5 multiple choice questions.
The true/false questions should be easy, just do 3x20 and you get 60.
ANSWER: 5
