Answer:

Step-by-step explanation:
According to the diagram
if
The center is o
Then Distance from O to B'C'=1
Distance from O to BC=1+1=2

<h3><u>More</u><u> </u><u>to</u><u> </u><u>know</u><u>:</u><u>-</u></h3><h3><u>Formulas</u><u> </u><u>related</u><u> </u><u>to</u><u> </u><u>surface</u><u> </u><u>area</u><u> </u><u>and</u><u> </u><u>volume</u><u>:</u><u>-</u></h3>

Answer:2
Step-by-step explanation:
x is located on the horizontal line, so when we look at 3 on the horizontal and trace that line up to where it hits the red line of the function you will find y, the vertical value which is 2
<h2>
Therefore x=2 and y= 1</h2>
Step-by-step explanation:
Given equations are
and
.........(2)
⇒
.......(1)
Equation (2)× 3 - Equation (1)

⇔
⇔-10y = -10
⇔y=1
Putting the value of y in equation (1) we get

⇔x =2
Therefore x=2 and y= 1
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: (m-1)(m-8)
Hope this helps