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Ne4ueva [31]
3 years ago
11

PLZ HELP !! Which of the points (s, t) is not one of the vertices of the shaded region of the

Mathematics
1 answer:
stira [4]3 years ago
3 0

Answer:

A(0,10)

Step-by-step explanation:

Given the 4 inequalities:

s ≥ 12 - 0.5t (1)

s ≥ 10 -t (2)

s ≤ 20-t (3)

s ≥ 0

t ≥ 0

Let analyse all 4 possible answer:

  • A(0, 10)

Let substitute this point into (1) we have: 10 ≥ 12 -0.5*0 Wrong

We do not choose A

  • B (0, 20)

Let substitute this point into (1) we have: 20 ≥ 12 -0.5*0  True

Let substitute this point into (2) we have: 20 ≥ 10 - 0 True

Let substitute this point into (3) we have: 20 ≤ 20 - 0 True

We choose B as the vertex

  • C. (4, 16)

Let substitute this point into (1) we have: 16 ≥ 12 -0.5*4  True

Let substitute this point into (2) we have: 16 ≥ 10 - 4 True

Let substitute this point into (3) we have: 16 ≤ 20 - 4 True

We choose C as the vertex

  • D. (16,4)

Let substitute this point into (1) we have: 4 ≥ 12 -0.5*16 True

Let substitute this point into (2) we have: 4 ≥ 10 - 16 True

Let substitute this point into (3) we have: 4 ≤ 20 - 16 True

We choose B as the vertex

Hence, the point A(0,10)  is not one of the vertices of the shaded region of the  set of inequalities.

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Answer:

a) P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

b) P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

c) P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=10, p=0.32)

The probability mass function for the Binomial distribution is given as:

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Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

Part b

P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]

P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211  

P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

Part c

P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264

P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218

P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123

P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

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