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:
C.
Step-by-step explanation:
(B. looks like it may also be one).
Answer:
-9x^9y^5
Step-by-step explanation:
Do the multiply first:
-16x^9y^5+7x^9y^5
=x^9y^5(-16+7)
=-9x^9y^5
<h2>
Answer:</h2><h3>f(4) = 122</h3>
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<h3>
Calculate</h3>
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<h3>Substitute</h3>
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<h3>Calculate the power</h3>
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<h3>Calculate the product or quotient</h3>
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<h3>Calculate the sum or difference</h3>
<em>I hope this helps you</em>
<em>:)</em>
It is 5/4 yards, you just add
When the denominator ( the bottom number ) is the same, just add the numerator ( the top number ) together :)
hope this help
