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oksian1 [2.3K]
3 years ago
14

a store received 500 containers of milk to be sold by Febuary 1. Each container the store sold $0.83 and sold for $1.67. The sto

re signed a contract with the distributor in which the distributor agreed to a $0.50 refund for every container not sold by Febuary 1. If 470 containers were sold bu February 1, how much profit did the store make?
Mathematics
1 answer:
FrozenT [24]3 years ago
6 0
$409.80.

The amount of net profit the store makes on each container is given by 1.67-0.83 (the amount it is sold for subtracted by the amount it costs the store), which is $0.84 per container.  They sell 470 containers, so the net profit at this point is 470(0.84) = $394.80.

However, since the distributor is giving the store a $0.50 refund on all every container under 500 that the store sells, the store gets additional money back:

500-470 = 30 containers not sold
30(0.50) = 15

So the total profit is $394.80 + 15 = $409.80.
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Answer:

a) P=0.3174

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c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}

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P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174

b)  What is the probability that at least one is essay?

P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232

c) What is the probability that two or more are essay?

P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594

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Hello from MrBillDoesMath!

Answer:

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Each term in   -3x + 6 has 3 as a factor so the factorization is

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