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Schach [20]
3 years ago
13

Choose the option that best answers the question.The points A(0, 0), B(0, 4a – 5) and C(2a + 1, 2a + 6) form a triangle. If Angl

e ABC = 90, what is the area of triangle ABC? 102 120 132 144 156
Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0

Answer:

The correct option is 1.

Step-by-step explanation:

Given information: The coordinates of a right angled triangle ABC are A(0, 0), B(0, 4a – 5) and C(2a + 1, 2a + 6). Angle ABC = 90°.

It means AB and BC are legs of the right angled triangle ABC.

Side AB lies on the y-axis because the x-coordinate of both A and B is 0.

Two legs are perpendicular to each other. So, BC must be parallel to x-axis and the y-coordinate of both B and C is must be same.

4a-5=2a+6

4a-2a=5+6

2a=11

Divide both sides by 2.

a=\frac{11}{2}

The value of a is 2. So the coordinates of triangle ABC are

B(0,4a-5)=B(0,4(\frac{11}{2})-5)\Rightarrow B(0,17)

C(2a+1,2a+6)=C(2(\frac{11}{2})+1,2(\frac{11}{2})+6)\Rightarrow C(12,17)

The area of a triangle is

Area=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|

The area of triangle ABC is

Area=\frac{1}{2}|0(17-17)+0(17-0)+12(0-17)|

Area=\frac{1}{2}|12(-17)|

Area=\frac{1}{2}|-204|

Area=\frac{1}{2}(204)

Area=102

The area of triangle ABC is 102. Therefore the correct option is 1.

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Answer:

a) We have the standard deviation and the mean, so it is appropriate to use the normal distribution to find probabilities for x.

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Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

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After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

A sample of size 45 will be drawn from a population with mean 53 and standard deviation 11.

This means that \mu = 53.

We have to find the standard deviation of the sample, that is:

\sigma = \frac{11}{\sqrt{45}} = 1.64

(a) Is it appropriate to use the normal distribution to find probabilities for x?

We have the standard deviation and the mean, so it is appropriate to use the normal distribution to find probabilities for x.

(b) Find the probability that x will be between 54 and 55.

This is the pvalue of the Z score when X = 55 subtracted by the pvalue of the Z score when X = 54.

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Z = 1.22

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Z = \frac{X - \mu}{\sigma}

Z = \frac{54 - 53}{1.64}

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(c) Find the 47th percentile of x. Round the answer to at least two decimal places.

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This is between Z = -0.07 and Z = -0.08. So we use Z = -0.075.

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3 years ago
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Answer:

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