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aleksandr82 [10.1K]
3 years ago
12

1 point) A computer retail store has 11 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the

retail store or the buyer, 3 of the computers in stock have defective hard drives. Assume that the computers are selected at random. (a) In how many different ways can the 3 computers be chosen
Mathematics
1 answer:
jekas [21]3 years ago
5 0

Answer: 165 ways

Step-by-step explanation:

The process of selection is done by using the combination formula for selection. If given "n" items and we are to choose "r" items from this given "n", the formula to use is denoted by:

nCr. = n! /(n-r)! × r!

Where

n! = n * (n-1) * (n-2)... *3*2*1.

In this case the number of items given = 11

And the number of items to be chosen = 3,

Hence the number of ways to do this = 11C3= 11!/(11-3)! × 3!

=165 ways.

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nevsk [136]

Answer: Option C)1 over 15 minus 1 over x equals 1 over 20

Explanation:

Since, Micah can fill a box with books in 15 minutes.

Therefore, the work done by Micah in one minute= 1/15

Also, Sydney takes the books out puts them on a shelf.

And the times taken by Micah when Sydney is also taking the books outside from the self= 20 minutes

Therefore, the work done by Micah in one minute when Sydney  taking books out of the box= 1/20

Let Sydney alone takes x minutes to take books outsides the shelf.

Then, work done by Sydney in one minute=1/x

Thus, the work done by Sydney( by taking books out of the box)=  the work done by Micah - work done by Micah and Sydney simultaneously= 1/15-1/20

⇒1/x=1/15-1/20

⇒1/15-1/20=1/x

⇒1/15-1/x=1/20 is the required expression.

Therefore, Option C is correct.



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