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zepelin [54]
3 years ago
7

claire mowed 5 lawns last week. show mowed each lawn in 7/12 hour. she mowed the same lawns this week in 5/12 hour each using he

r new lawn mower. How many times longer was Claire's time to now all the lawns last week than this week?
Mathematics
1 answer:
Zinaida [17]3 years ago
5 0

Answer:

\dfrac{5}{6} hour or 50 minutes

Step-by-step explanation:

<u>Last week:</u>

Claire mowed 5 lawns

She mowed each lawn in \frac{7}{12} hour.

She spent

5\cdot \dfrac{7}{12}=\dfrac{35}{12}=2\dfrac{11}{12}

hours to mow 5 lawns.

<u>This week:</u>

Claire mowed 5 lawns

She mowed each lawn in \frac{5}{12} hour.

She spent

5\cdot \dfrac{5}{12}=\dfrac{25}{12}=2\dfrac{1}{12}

hours to mow 5 lawns.

<u>Difference:</u>

2\dfrac{11}{12}-2\dfrac{1}{12}=\dfrac{11}{12}-\dfrac{1}{12}=\dfrac{10}{12}=\dfrac{5}{6}

hour or 50 minutes.

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According to a study, 50 % of adult smokers started smoking before 21 years old. 5 smokers 21 years old or older are randomly se
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Answer:

a) The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

Step-by-step explanation:

For each smoker, there are only two possible outcomes. Either they started smoking before 21 years old, or they did not. Smokers are independent, which means that the binomial probability distribution is used to solve this question.

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 combinations 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.

50% of adult smokers started smoking before 21 years old.

This means that p = 0.5

5 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.

This means that n = 5.

a) The probability that at least 2 of them started smoking before 21 years of age is

This is:

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

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

P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125

P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625

P(X < 2) = P(X = 0) + P(X = 1) = 0.03125 + 0.15625 = 0.1875

The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is

This is:

P(X \leq 4) = 1 - P(X = 5)

In which

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

P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125

P(X \leq 4) = 1 - P(X = 5) = 1 - 0.03125 = 0.96875

The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is

This is P(X = 3). So

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

P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125

The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

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