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Rufina [12.5K]
4 years ago
13

Section J in an arena has 20 rows each row has 15 seats all tickets cost $18 each people if all the seats are sold how much mone

y can I collect for Section J
Mathematics
2 answers:
Novosadov [1.4K]4 years ago
5 0
First do 20×15, then the answer multiply by 18.
Arturiano [62]4 years ago
4 0
20 x 15 = 300 and 300 x 18 = $5,400 so try that
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Mary Palm's checking account had a starting balance of $785.63. She wrote a check for $57.00 for groceries and a check for $125.
wariber [46]
785.63-(57+125)
=785.63-182
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5 0
4 years ago
NEED ANSWER QUICK!!
dybincka [34]

Answer:

$192

Step-by-step explanation:

2400/48=50

50x3.84=192

3 0
3 years ago
At Glenridge High School, 20 percent of the
Makovka662 [10]

Answer:

40 percent

Step-by-step explanation:

60 - 20 = 40

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Hope this helps!

3 0
3 years ago
12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests
Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem, we have that:

12% of children are nearsighted. This means that p = 0.12.

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so n = 18

This probability is:

P = P(X = 0) + P(X = 1)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{18,0}.(0.12)^{0}.(0.88)^{18} = 0.1002

P(X = 1) = C_{18,1}.(0.12)^{1}.(0.88)^{17} = 0.2458

So

P = P(X = 0) + P(X = 1) = 0.1002 + 0.2458 = 0.3460

There is a 34.60% probability that 0 or 1 of them is nearsighted.

5 0
3 years ago
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gizmo_the_mogwai [7]

Answer:

4

Step-by-step explanation:

on edg

5 0
4 years ago
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