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3 years ago

Uses fewer than 5 coins to pay 60 c

Mathematics

2 answers:

postnew [5]3 years ago

7 0

It would be two quarters and one dime

MaRussiya [10]3 years ago

3 0

2 quarters and 1 dime
25+25+10=60

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A local movie theater charges $12 for an adult ticket and $10 for a child's ticket. a group of eight people spent a total of $86

oksian1 [2.3K]

3 adults and 5 kids.

You can get this by setting up the following system of equations.

x + y = 8 (number of people)
10x + 12y = 86 (cost)

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3 years ago

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4 years ago

the accountant charged $35 for the first hour of work and $23 for each hour he did after that he earned a totL or $127 how many

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3 years ago

Read 2 more answers

A company is producing two types of ski goggles. Thirty percent of the production is of type A, and the rest is of type B. Five

exis [7]

Answer:

$\dfrac{14}{29}$

Step-by-step explanation:

Let P(A) be the probability that goggle of type A is manufactured

P(B) be the probability that goggle of type B is manufactured

P(E) be the probability that a goggle is returned within 10 days of its purchase.

According to the question,

P(A) = 30%

P(B) = 70%

P(E/A) is the probability that a goggle is returned within 10 days of its purchase given that it was of type A.

P(E/B) is the probability that a goggle is returned within 10 days of its purchase given that it was of type B.

$P(A \cap E)$ will be the probability that a goggle is of type A and is returned within 10 days of its purchase.

$P(B \cap E)$ will be the probability that a goggle is of type B and is returned within 10 days of its purchase.

$P(E \cap A) = P(A) \times P(E/A)$

$P(E \cap A) = \dfrac{30}{100} \times \dfrac{5}{100}\\\Rightarrow P(E \cap A) = 1.5 \%$

$P(E \cap B) = P(B) \times P(E/B)$

$P(E \cap B) = \dfrac{70}{100} \times \dfrac{2}{100}\\\Rightarrow P(E \cap B) = 1.4 \%$

$P(E) = 1.5 \% + 1.4 \% \\P(E) = 2.9\%$

If a goggle is returned within 10 days of its purchase, probability that it was of type B:

$P(B/E) = \dfrac{P(E \cap B)}{P(E)}$

$\Rightarrow \dfrac{1.4 \%}{2.9\%}\\\Rightarrow \dfrac{14}{29}$

So, the required probability is $\dfrac{14}{29}.$

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4 years ago

How is it going to tell me the answer

Anon25 [30]

I dont know man you gotta ask a question

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3 years ago