Answer:
30 cans
Step-by-step explanation:
We can use ratios
15 cans x cans
---------------- = ------------------
7*2 people days 4*7 people days
Using cross products
15 * 28 = 14 * x
Divide by 14
15 * 28/14 = x
30 =x
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:
<h2>20 ft, 48 ft, 52 ft</h2>
Step-by-step explanation:
The side lengths of ΔEFG are:
5 ft, 12 ft and 13 ft.
The scale factor: s = 4
All sides of the ΔKLM are four times longer than the sides of the ΔKLM.
Therefore the side lengths of KLM are:
5 ft · 4 = 20 ft
12 ft · 4 = 48 ft
13 ft · 4 = 52 ft
