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3 years ago

If there are four teams in a league, how many games will have to be played so that each team plays every other team once?

Mathematics

1 answer:

Sedaia [141]3 years ago

3 0

Answer:

6 matches

Step-by-step explanation:

Let’s call the teams A B C D

A will play B C D = 3 matches

B will play only C and D as it already played A, making 2 matches

C will play D, making 1 match

D has already played all

Total number of matches is thus 3 + 2 + 1 = 6 matches

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3 years ago

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A box of Munchkins contains chocolate and glazed donut holes. If Jacob ate 2 chocolate

neonofarm [45]

Answer:

24 munchkins.

Step-by-step explanation:

Let C be the number of chocolate and D be number of glazed donut holes in the original box.

We are told if Jacob ate 2 chocolate munchkins, then 1/11 of the remaining Munchkins would be chocolate. We can represent this information as:

$C-2=\frac{1}{11}*(C+D-2)...(1)$

We are also told if he instead added 4 glazed Munchkins to the original box, 1/7 of the Munchkins would be chocolate. We can represent this information as:

$C=\frac{1}{7}*(C+D+4)...(2)$

Upon substituting C's value from equation (2) in equation (1) we will get,

$\frac{1}{7}*(C+D+4)-2=\frac{1}{11}*(C+D-2)$

Let us have a common denominator on right side of equation.

$\frac{1}{7}*(C+D+4)-\frac{7*2}{7}=\frac{1}{11}*(C+D-2)$

$\frac{C+D+4-14}{7}=\frac{1}{11}*(C+D-2)$

Multiplying both sides of our equation by 7, we will get,

$7*\frac{C+D-10}{7}=7*\frac{1}{11}*(C+D-2)$

$C+D-10=\frac{7}{11}*(C+D-2)$

Multiplying both sides of our equation by 11, we will get,

$11*(C+D-10)=11*\frac{7}{11}*(C+D-2)$

$11*(C+D-10)=7*(C+D-2)$

$11C+11D-110=7C+7D-14$

$11C-7C+11D-7D=-14+110$

$4C+4D=96$

$4(C+D)=96$

$(C+D)=\frac{96}{4}$

$(C+D)=24$

Therefore, the total number of Munchkins in original box is 24.

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3 years ago

Complete the following statement.<br><br> 16 fl oz = pt

Elden [556K]

Answer:

Step-by-step explanation:

divide the volume value by 16

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3 years ago