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12345 [234]
2 years ago
5

Can someone please help me?! Thanks

Mathematics
2 answers:
omeli [17]2 years ago
5 0

Answer:

answer: A: x \leq 48

Step-by-step explanation:

\frac{x}{12} + 3 \leq 7

Subtract 3 from both sides of the inequality:

\frac{x}{12} + 3 - 3 \leq 7 - 3

\frac{x}{12}  \leq 4

Multiply both sides of the inequality by 12 to solve for x:

\frac{x}{12} (\frac{12}{1})  \leq 4 (12)

x \leq 48

Mekhanik [1.2K]2 years ago
3 0

Answer:

maybe b option

have a great day

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Answer:

a) P=0.1721

b) P=0.3528

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Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

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This probability can be calculated as:

P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

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P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}

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P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}

Then we have

P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453

The sum of the probabilities is

P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981

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