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Taya2010 [7]
3 years ago
12

Guys please help me I don’t understand

Mathematics
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer:

22

Step-by-step explanation:

6x+3=7x-19

6x-7x=-19-3

-x=-22 |÷(-1)

x=22

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A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 y
Anestetic [448]

Answer:

\frac{1}{4180}

Step-by-step explanation:

GIVEN: A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement.

TO FIND: What is the probability that the first two candies drawn are orange and the third is green.

SOLUTION:

Total  candies in the bag =57

Probability that first ball is orange, P(A)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}

                                                       =\frac{3}{57}=\frac{1}{19}

Probability that second ball is orange, P(B)=\frac{\text{total orange candies in bag}}{\text{total candies in bag}}

                                                        =\frac{2}{56}=\frac{1}{28}

Probability that third ball is green, P(C)=\frac{\text{total green candies in bag}}{\text{total candies in bag}}

                                                                 =\frac{7}{55}

Now, probability that first two balls are orange and third is green is

=P(A)\times P(B)\times P(C)

=\frac{1}{19}\times\frac{1}{28}\times\frac{7}{55}

=\frac{1}{19}\times\frac{1}{4}\times\frac{1}{55}

=\frac{1}{4180}

Hence,  probability that first two balls are orange and third is green is \frac{1}{4180}

3 0
4 years ago
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