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Lubov Fominskaja [6]
3 years ago
15

one container holds a 1 7/8 quarts of water and a second container holds 5 3/4 quarts of water. how many more quarts of water do

es the second container hold than the first container?
Mathematics
1 answer:
dimulka [17.4K]3 years ago
8 0
So lets solve this!

Quarts of water 1st container holds= 17/8

Quarts of water 2nd container holds= 5 3/4

Quarts of water does the second container holds than the first container=?

First of all we need to change 5 3/4 into an improper fraction

5 3/4= 23/4

Secondly, we need to find the LCM of 17/8 and 23/4

LCM:

17/8= 51/24

23/4= 138/24

So now we have got both of the LCM's 

Quarts of second container holds the first container= 23/4 - 138/24=87/24

Your answer is 87/24

Hope this Helps!




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If you roll two six-faced dice together, you will get 36 possible outcomes. 4 pts 1. List all possible outcomes of the experimen
bezimeni [28]

Given:

Two dice are rolled together.

Total number of possible outcomes.

To find:

The list of total possible outcomes.

The probability of getting a sum of 11 in these outcomes.

The probability of getting a sum less than or equal to 4.

The probability of getting a sum of 13 or more.

Solution:

If two dice are rolled together, then the total number of possible outcomes is 36 and list of total possible outcomes is

S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Sum of 11 in these outcomes = {(5,6),(6,5),(6,6)} = 3

The probability of getting a sum of 11 in these outcomes is

P(\text{sum=11})=\dfrac{3}{36}

P(\text{sum=11})=\dfrac{1}{12}

Therefore, the probability of getting a sum of 11 in these outcomes is \dfrac{1}{12}.

Sum less than or equal to 4 = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} = 6

The probability of getting a sum less than or equal to 4 is

P({sum\leq 4})=\dfrac{6}{36}

P({sum\leq 4})=\dfrac{1}{6}

Therefore, the probability of getting a sum less than or equal to 4 is \dfrac{1}{6}.

Sum of 13 or more = empty set because maximum sum is 12.

The probability of getting a sum of 13 or more is

P(sum\geq 13)=\dfrac{0}{36}

P({sum\geq 13})=0

Therefore, the probability of getting a sum of 13 or more is 0.

8 0
3 years ago
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