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lisov135 [29]
3 years ago
11

How do you use the least common multiple to write two or more fractions with a common denominator

Mathematics
1 answer:
Triss [41]3 years ago
4 0
First you compare the denomonators and see if they have any common multiples, then what you do to one side you do to the other so you multiply so you get the same number on both denomonators.finally you multipybthe same number you did ti he botto. to the top.p.s. your welcome
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Solve for x: x3 = 512 Your answer may be a whole number or a fraction.
Likurg_2 [28]

Answer:

x=8

Step-by-Step Explanation:

View provided picture

7 0
3 years ago
The domaln of the following relation R ((6, -2), (1, 2), (-3, -4), (-3, 2)} Is (1 polnt)
castortr0y [4]

Answer:

A. -3,-3,1,6

Step-by-step explanation:

The domain of a relation is the x-coordinates.

7 0
3 years ago
According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are re
Luda [366]

Answer:

The questions asked are

If you randomly select 4 peanuts

1. Compute the probability that exactly three of the four M&M’s are brown

2. Compute the probability that two or three of the four M&M’s are brown.

3. Compute the probability that at most three of the four M&M’s are brown.

4. Compute the probability that at least three of the four M&M’s are brown.

Step-by-step explanation:

Given the following information

Brown=12%. P(B)=0.12

Yellow=15%. P(Y)=0.15

Red=12%. P(R), =0.12

Blue=23%. P(B) =0.23

Orange, =23%. P(O) =0.23

Green=15%. P(G)=0.15

Question 1.

They are independent events

If there are exactly three brown and the last is not brown

P(B n B n B n B')

P(B)×P(B)×P(B)×P(B')

0.12×0.12×0.12×(1-P(B))

0.001728×(1-0.12)

0.001728×0.88

0.00152.

0.152%

2. If two or three are brown

I.e we are going to two brown and two none brown or three brown and one not brown. (P(B)×P(B)×P(B')×P(B'))+ (P(B)×P(B)×P(B'))

(0.12×0.12×0.88×0.88)+(0.12×0.12×0.12×0.88)

0.0112+0.00152

0.0127

1.27%

3. At most three brown out of four then we are going to have

BBBB', BBB'B', BB'B'B', B'B'B'B'

These are the cases of at most three brown.

P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B')×P(B') + P(B)×P(B')×P(B')×PB')+ P(B')×P(B')×P(B')×P(B')=

0.12×0.12×0.12×0.88+ 0.12×0.12×0.88×0.88+ 0.12×0.88×0.88×0.88+ 0.88×0.88×0.88×0.88=0.694

0.694

69.4%

4. At least 3 brown out of four selection

I.e BBBB', BBBB

These are the two options

P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B)×P(B)=

0.12×0.12×0.12×0.88 + 0.12×0.12×0.12×0.12

0.001728

0.173%

4 0
3 years ago
Read 2 more answers
What is the equation of a circle that has a diameter of 10 units if its center is at (3,8)?
astraxan [27]

Answer:

Step-by-step explanation:

(x-3)^2+(y-8)^2 = 25

6 0
3 years ago
Nendndnfbffjfjfjfjfjf
Whitepunk [10]

Answer:

a or c would be your answer hope this helps

Step-by-step explanation:

3 0
3 years ago
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