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3 years ago

Find the area of the triangle.

Mathematics

1 answer:

BabaBlast [244]3 years ago

5 0

Answer:

17.7 cm^2

Step-by-step explanation:

Use trig to find the height of the triangle. Then the area is bh/2.

Extend side BC to the right until it is vertically below point A. Draw a segment from point A vertically down until it intersects the extension of side BC. Call the point of intersection D. <D is a right angle.

Use triangle ABD to find the height, AD, of triangle ABC.

For <B of 37 deg, AD is the opposite leg. AB is the hypotenuse. The trig ratio that relates the opposite lefg to the hypotenuse is the sine.

sin B = opp/hyp

sin 37 deg = AD/13.1

AD = 13.1 * sin 37 deg

AD = 7.9

AD is the height of triangle ABC. BC is the base. We can find the area of triangle ABC.

area = bh/2

area = (4.5 cm)(7.9 cm)/2

area = 17.7 cm^2

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Step-by-step explanation:

The key to solving word (story) problems is to learn how to translate words and phrases into concise, precise math expressions (variables and operations). Make a table and add to it as you do problems:

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3 years ago

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

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b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

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A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$

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By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

The standard deviation of the sample mean weight is 0.123.

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This is the pvalue of Z when $X = 20.27$ .

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This is the value of $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

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$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

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2 years ago