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Anna71 [15]
2 years ago
14

1.2n+12!/n-1! 2. 4n+3!/2n+1!

Mathematics
1 answer:
Naddik [55]2 years ago
3 0
Are you sure there is no bracket in the denominators (n-1)! if not then the answer is:

1) 2n + 12!/2n -1 ===> 2n + (479,001,600)/n -1 or (2n²-n+12!)/n
2) 4n+3!/2n +1! ====>4n +3/n +1
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Answer:

a) P(R

And we can find this probability using the normal standard table or excel and we got:

P(z

b) P(R>110)=P(\frac{R-\mu}{\sigma}>\frac{110-\mu}{\sigma})=P(Z>\frac{110-100}{3.606})=P(Z>2.774)

And we can find this probability using the complement rule and the normal standard table or excel and we got:

P(z>2.774)=1-P(Z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the time for the step 1 and Y the time for the step 2, we define the random variable R= X+Y for the total time and the distribution for R assuming independence between X and Y is:

R \sim N(40+60 = 100,\sqrt{2^2 +3^2}= 3.606 s)  

Where \mu=65.5 and \sigma=2.6

We are interested on this probability

P(R

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{R-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(R

And we can find this probability using the normal standard table or excel and we got:

P(z

Part b

P(R>110)=P(\frac{R-\mu}{\sigma}>\frac{110-\mu}{\sigma})=P(Z>\frac{110-100}{3.606})=P(Z>2.774)

And we can find this probability using the complement rule and the normal standard table or excel and we got:

P(z>2.774)=1-P(Z

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Step-by-step explanation:

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