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STatiana [176]
1 year ago
13

Grayson needs to order some new supplies for the restaurant where he works. The restaurant needs at least 761 forks. There are c

urrently 205 forks. If each set on sale contains 10 forks, which inequality can be used to determine x, the minimum number of sets of forks Grayson should buy?options are.1. 761 ≥ 10(205+x)2. 761 ≤ 10(205+x)3. 761 ≥ 10x+2054. 761 ≤ 10x+205
Mathematics
1 answer:
Triss [41]1 year ago
7 0

The restaurant needs at least 761 forks.

There are currently 205 forks

Each set on sale contains 10 forks,

The number of set taht have to buy are x

Number of forks in x set are = 10x

Since, we need at least 761

So 761 should be graeter than equal to the sum of reamining forks and the new forks

i.e. 761 ≥ 10x + 205

Answer : 3. 761 ≥ 10x + 205

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Answer:

a) For this case the random variable X follows a hypergometric distribution.

b) E(X)= n\frac{M}{N}=10 \frac{5}{25}=2

Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1

c) P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565

d) P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474

Step-by-step explanation:

The hypergometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:

P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}

Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes

The expected value and variance for this distribution are given by:

E(X)= n\frac{M}{N}

Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}

a. What is the distribution of X?

For this case the random variable X follows a hypergometric distribution.

b. Compute the values for E(X) and Var(X)

For this case n=10, M=5, N=25, so then we can replace into the formulas like this:

E(X)= n\frac{M}{N}=10 \frac{5}{25}=2

Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1

c. What is the probability that none of the animals in the second sample are tagged?

So for this case we want this probability:

P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565

d. What is the probability that all of the animals in the second sample are tagged?

So for this case we want this probability:

P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474

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