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mr_godi [17]
3 years ago
13

The number of pieces of popcorn in a large movie theatre popcorn bucket is normally distributed, with a mean of 1515 and a stand

ard deviation of 15. Approximately what percentage of buckets contain between 1470 and 1560 pieces of popcorn?
Mathematics
1 answer:
Murrr4er [49]3 years ago
3 0

Answer: 99.7%, Im pretty confident about that. Im taking the test right now.

Step-by-step explanation:

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-1

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1) The height of the cylinder is equal to the diameter of the sphere.

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At the movie theatre, child admission is and adult admission is . On Wednesday, tickets were sold for a total sales of . How man
kirill [66]

The child tickets were sold that day is 82.

Given the cost of child admission $6.30 and the cost of adult admission $9.80 and on Wednesday 144 tickets were sold for a total sales of $1124.20.

Let the child tickets are x and adult tickets are y.

On Monday, 144 tickets were sold for a total sales.

x+y=144              ........(1)

Total sales = $1124.20

Need to find the child tickets sold on Wednesday.

Total sales = 6.30x+9.80y

Total sales = $1124.20

So,6.30x+9.80y=1124.20                        .......(2)

Find the value of y from equation (1), we get

x+y-x=144-x

y=144-x

Now, we will substitute the value of y in equation (2) to find the value of x, we get

6.30x+9.80(144-x)=1124.20

Apply the distributive property that is a(b+c)=ab+ac, we get

6.30x+9.80×144-9.80x=1124.20

6.30x+1411.2-9.80x=1124.20

Combine the variable terms on left side, we get

1411.2-3.5x=1124.20

subtract 1411.2 from both sides, we get

1411.2-3.5x-1411.2=1124.20-1411.2

-3.5x=-287

Divide both sides with -3.5x, we get

-3.5x/(-3.5)=-287/(-3.5)

x=82

Therefore, 82 child tickets were sold on Wednesday when cost of child admission $6.30 and the cost of adult admission $9.80.

Learn more about variable from here brainly.com/question/14786534

#SPJ4

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