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DENIUS [597]
2 years ago
13

Classify the triangle. obtuse equiangular right acute

Mathematics
2 answers:
wlad13 [49]2 years ago
6 0

Answer:

Acute

Step-by-step explanation:

Acute=less than 90 degrees

Obtuse=more than 90 degrees

Right=1 side has 90 degrees

Equiangular=every side has the same degree

Therefore, it would be Acute

CaHeK987 [17]2 years ago
3 0

Answer:

acute triangle

Step-by-step explanation:

We have 3 angles that are not equal to each other

They are all smaller than 90 degrees which means they are acute angles

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What is the slope of the line NP
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Answer:

The slope of NP is \frac{-5}{7} ⇒ B

Step-by-step explanation:

  • In any square, every two opposite sides are parallel and every two consecutive sides are perpendicular
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  • The product of the slopes of the perpendicular lines is -1, which means if the slope of one m, then the slope of the other is \frac{-1}{m} (reciprocal m and change its sign)
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In square MNPR

∵ MN and NP are adjacent sides

∴ MN ⊥ NP

∵ M = (3, 8) and N = (-2, 1)

∴ x1 = 3 and y1 = 8

∴ x2 = -2 and y2 = 1

→ Use the rule of the slope above to find the slope of MN

∴ m(MN) = \frac{1-8}{-2-3}

∴ m(MN) = \frac{-7}{-5}

∴ m(MN) = \frac{7}{5}

∵ MN ⊥ NP

∴ The product of their slopes = -1

→ Reciprocal the slope of MN and change its sign

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3 years ago
Assume that the probability of a defective computer component is 0.02. Components arerandomly selected. Find the probability tha
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Answer:

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.

Step-by-step explanation:

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested.

First six not defective, each with 0.98 probability.

7th defective, with 0.02 probability. So

p = (0.98)^6*0.02 = 0.0177

0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.

Find the expected number and variance of the number of components tested before a defective component is found.

Inverse binomial distribution, with p = 0.02

Expected number before 1 defective(n = 1). So

E = \frac{n}{p} = \frac{1}{0.02} = 50

Variance is:

V = \frac{np}{(1-p)^2} = \frac{0.02}{(1-0.02)^2} = 0.0208

The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.

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