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 = - 1 ± i 
Step-by-step explanation:
Given
3x² = - 12 - 6x ( add 6x to both sides )
3x² + 6x = - 12 ( divide through by 3 )
x² + 2x = - 4
To complete the square
add ( half the coefficient of the x- term )² to both sides
x² + 2(1)x + 1 = - 4 + 1
(x + 1)² = - 3 ( take the square root of both sides )
x + 1 = ± 
= ± i ( subtract 1 from both sides )
x = - 1 ± i
Answer:
D)
Step-by-step explanation:
Answer:
5x-7
Step-by-step explanation:
= -8x - 12 - 3x + 5
= -8x - 3x - 12 + 5
= 5x-7
Answer:
X< 7/12
Step-by-step explanation:
