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olga2289 [7]
3 years ago
12

Bryan is setting charis in rows for graduation ceremony he has 50 black chars,and 60 white charis each row will have the same nu

mber of chairs what is the greatest number of chairs that he can fit in each row?how many rows of each coler chair will there be
please help and show me how you did it thanks!!
Mathematics
2 answers:
konstantin123 [22]3 years ago
5 0
Assuming that no row can have a combination of black and white chairs, there will be 11 rows, each consisting of 10 chairs. 5 of these rows will have black chairs, and 6 of these rows will have white chairs. I arrived at this by finding the GCF (or greatest common multiple) of 50 and 60, which is 10. That number represents the maximum number of chairs per row. Then, I added 50 and 60, to get 110, and divided that by the number of chairs per row, 10, to arrive at there being 11 rows. 50/110 chairs are black, meaning 5 rows are black, and 60/110 chairs are white, meaning 6 rows are white. Hope this helps! <em>~ArchimedesEleven</em>
Ronch [10]3 years ago
3 0
The factors of 60 are 1,2,3,5,6,10,12,20,30,and 60 
the factors of 50 are 1,2,5,10,15,and 30
so the GCF is 10
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Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml
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<h2>Answer:</h2>

a)

The probability that both televisions work is:  0.42

b)

The probability at least one of the two televisions does not​ work is:

                          0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

          9-3=6

a)

The probability that both televisions work is calculated by:

=\dfrac{6_C_2}{9_C_2}

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42

b)

The probability at least one of the two televisions does not​ work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5

and the probability both do not work is:

=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833

Hence, Probability that atleast does not work is:

             0.5+0.0833=0.5833

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You have to turn 65 into a decimal. That would be .65
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