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tekilochka [14]
3 years ago
15

But The aquarium charges $12 per ticket for adults and $5per ticket for children. A group of 90 children and adult chaperones vi

sited the aquartum on a held trip. If the total cost of their tickets was $548, how mamy chaperones were there?
Mathematics
1 answer:
White raven [17]3 years ago
7 0

Answer:

There were 76 childrens and 14 adults.

Step-by-step explanation:

Since the group has a total of 90 children and adults, then the sum of the number of adults with the number of children must be equal to 90 as shown below:

children + adults = 90

Since the total cost for their tickets was 548 then the number of children multiplied by the price of their ticket summed by the number of adults multiplied by the price of their ticket must be equal to that. We have:

5*children + 12*adults = 548

With these two equations we have a system of equations shown below:

children + adults = 90

5*children + 12*adults = 548

In order to solve this we will multiply the first equation by -5, and sum both equations we have:

-5*children - 5*adults = -450

5*children + 12*adults = 548

7*adults = 98

adults = 98/7 = 14

children + 14 = 90

children = 90 - 14 = 76

There were 76 childrens and 14 adults.

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Exponents are done before addition, subtraction, multiplicaton, and division

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2 years ago
Determine the surface area of the cylinder below.
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Step-by-step explanation:

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3 years ago
Can anyone help me with this. Click to see
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8 0
2 years ago
2. A statistics student plans to use a TI-84 Plus calculator on her final exam. From past experience, she estimates that there i
Anarel [89]

Answer:

  1. P(≥1 working) = 0.9936
  2. She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.

Step-by-step explanation:

1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...

... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936

2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.

If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.

This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.

_____

My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)

7 0
3 years ago
What is x?
iren2701 [21]
X = 5 - 

-7x  - 3x + 2 = -8x - 8
+ 7x                 +7x
-3x + 2 = -x - 8
+3x        +3x
2 = 2x -8
+8       +8
10 = 2x
/2      /2
x = 5
7 0
3 years ago
Read 2 more answers
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