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icang [17]
3 years ago
6

Cora has 1/2 hour to do 5chores. She palns to spend the same fraction of an hour on each chore. She wants to use the number line

to help her determine what fraction of an hour she can spend on each chore
Mathematics
2 answers:
kiruha [24]3 years ago
4 0
Hi! So to find this answer, you could do it two ways. Firstly, you could spend a major amount of time doing this in your head. Or second, you could use a calculator and do 30/5. This would give you 6. So 6 minutes on each chore.
Free_Kalibri [48]3 years ago
4 0
If you want to use a number line create a number line going up to 30 (1/2 an hour). Then, divide it by 5 and then you can see that each section is 6 units.
So, she spends 6 min. on each chore.
6/60=1/10
She spends 1/10 of an hour on each chore.

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Answer:

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

Step-by-step explanation:

For sampling with replacement, we use the binomial distribution. Without replacement, we use the hypergeometric 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 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.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

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)!}

Sampling with replacement:

I consider a success choosing a black ball, so p = \frac{50}{150+50} = \frac{50}{200} = 0.25

We want 2 black balls and 2 white, 2 + 2 = 4, so n = 4, and we want P(X = 2).

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

P(X = 2) = C_{4,2}.(0.25)^{2}.(0.75)^{2} = 0.2109

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Sampling without replacement:

150 + 50 = 200 total balls, so N = 200

Sample of 4, so n = 4

50 are black, so k = 50

We want P(X = 2).

P(X = 2) = h(2,200,4,50) = \frac{C_{50,2}*C_{150,2}}{C_{200,4}} = 0.2116

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

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