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.
1. 550 pages left
55×4=220+330=550
2. 10 hours
330÷55=6+4=10
Answer:
D
Step-by-step explanation:
Let x be Bill’s scores.
Let y be Theo’s scores.
x + y = 90
x² = y
Solve for x in the second equation.
x = √y
Put x as √y in the first equation and solve for y.
√y + y = 90
√y = 90 - y
y = (90 - y)²
y = - y² - 180y - 8100
0 = -y² - 181y - 8100
0 = (-y+81)(y-100)
-y+81=0
y = 81
y - 100 = 0
y = 100
100 does not work if we plug in to check, so y is 81.
y = 81
Put y as 81 in the first equation and solve for x.
x + 81 = 90
x = 9
Bill scored 9 goals and Theo scored 81.
Answer:
(2,-1) and 5
Step-by-step explanation:
It’s A. I hope that helps
