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ioda
3 years ago
6

Perimeter of quadrilateral

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
3 0

Answer:

The answer to your question is:  14 + 2√40  = 26.6 units

Step-by-step explanation:

Data

A ( -5, 4)     B (-3, -2)     C (4, -2)     D (2, 4)

Formula

d = √(x2 - x1)² + (y2 - y1)²

Perimeter = dAB + dBC + dCD + dAD

Process

dAB = √(-3 + 5)² + (-2 - 4)²

dAB = √(2)² + (-6)²

dAB = √4 + 36

dAB = √40 units

dBC = √(4 + 3)² + (-2 + 2)²

dBC = √(7)²

dBC = √49

dBC = 7 units

dCD = √(2 - 4)² + (4 + 2)²

dCD = √(2)² + (6)²

dCD = √40 units

dAD = √(2 + 5)² + (4 - 4)²

dAD = √49

dAD = 7 units

Perimeter = √40 + 7 + √40 + 7

Perimeter = 14 + 2√40  = 26.6 units

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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.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

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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.

We apply the same formula but as a sum:

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

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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}

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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

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

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