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user100 [1]
3 years ago
5

At a baseball game, a vender sold a combined total of 170 sodas and hot dogs. the number of hot dogs sold was 50 less than the n

umber of sodas sold. find the number of sodas sold and the number of hot dogs sold
Mathematics
1 answer:
Juliette [100K]3 years ago
8 0

Let the number of sodas sold be x and the number of hotdogs sold be y. We can assemble a system of equations from the information given and solve.


x+y=170 (We know that soda + dogs is 170)

x-50=y (We know that dogs is equal to 50 less than sodas)


We can use substitution to solve this system of equations.

x+y=170

Sub in the value of y from the second equation

x+(x-50)=170

2x-50=170

2x=220

x=110


We now know the number of sodas sold was 110. Lets now plug this value into the equation to solve for y.

x+y=170

110+y=170

y=60


So, there were 60 hotdogs sold and 110 sodas sold.

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The length of a rectangle is 3 inches longer than its width. If the perimeter of the rectangle is 54 inches, find it’s length an
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Answer: Length = 15, Width = 12

Step-by-step explanation:

L = length

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The scuba club has 20 members, 5 girls and 15 boys. What is the ratio of girls to boys in the scuba club?
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3 years ago
7) PG & E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many
BabaBlast [244]

Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

============================================================

Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

-----------------

Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\

Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.

-----------------

<h3>Answer: 495</h3>
5 0
2 years ago
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