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2 years ago

How can they redistribute the money so that each child has the same amount? Check all that apply.

1 answer:

8 0

Depending on how much money it is and the children for an example if it’s 4 children and $40 do 40 divided by 4 = $10 a child

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Determine the Value of x,y and z.

Aleksandr-060686 [28]

Answer:

x=9

y=16

z=12

Step-by-step explanation:

7 0

3 years ago

C. Read and analyze the problem, then solve.

Sliva [168]

Given parameters:

Total length of curtains bought = 20m

She used 8m for the living room

She used 7m for the dining room

unknown:

Quantity used for the kitchen = ?

Solution:

Quantity used for the kitchen = Total length - (living room + dining room)

Input the parameters given solve;

Quantity used for the kitchen = 20 - (8 + 7) = 5m

The length of curtain used for the kitchen is 5m

5 0

3 years ago

Finding area and perimeter for each shape

san4es73 [151]

Answer: first one is 15.01

Step-by-step explanation:

7 0

3 years ago

This is middle school work

inysia [295]

Answer:

The deer population decreased. The producers increased. So the number of different species increased.

Step-by-step explanation:

wolves are reintroduced ---> deer population decreases --> producer population begins to increase --> other species numbers increase as well

7 0

2 years ago

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

3 years ago