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Anni [7]
3 years ago
8

Sophie and Simon are peeling a pile of potatoes for lunch in the cafeteria. Sophie can peel all the potatoes by herself in 45 mi

nutes, while it would take Simon 30 minutes to do the job working alone. If Sophie and Simon work together to peel the potatoes, how long will it take them?
Mathematics
2 answers:
natima [27]3 years ago
5 0

Answer:

18 Minutes

Step-by-step explanation:

Simon and sophie each do a fraction of the job per minute like this:

-Simon does \frac{1}{30} per minute (if we multiply this by 30 we have the completed job)

-Sophie does \frac{1}{45} of the job per minute (again if we multiply this by 45 we have the completed job)

We will add this two to know how much of the job they do together in a minute:

\frac{1}{30}+\frac{1}{45}=\frac{1}{18}

They will do \frac{1}{18} of the job per minute.

So the total time will be when they do \frac{18}{18}=1, we can put this in one equation like this:

\frac{1}{18}x=1

where x is the total times we need to do\frac{1}{18} of the job to complete it. We solve for x:

x=18minutes

Zanzabum3 years ago
4 0

Answer:

18 Minutes

Step-by-step explanation:

I did the test

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vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

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Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

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P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007

c) In this case, the distribution is B(1200,0.01)

P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366

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