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3 years ago

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Determine whether these triangles are similar. If so, circle the appropriate reason below. Please show all work.

1 answer:

uysha [10]3 years ago

3 0

Answer:

Yes these triangles ate similar.

Step-by-step explanation:

The two triangles are similar if their third angle is equal. The sum of all angles of triangles must be equal to 180 degrees. The triangle 1 has sum of 180 degrees and triangle 2 has sum of 180 degrees. These triangles have third angle same which is 45 degrees.

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3 years ago

In class, Matthew stated, "I think that a negative is like an opposite. That is why multiplying a negative times a negative equa

Black_prince [1.1K]

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3 years ago

Literal Equations: 8c + 6j = 5p, for c

bixtya [17]

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3 years ago

BD is the angle bisector of ZABC.

mojhsa [17]

Answer:

B:25°

Step-by-step explanation:

Now $\angle ABC$ gets bisected by $BD$ .

$\angle ABD=25^{\circ}$

Since, the angle is bisected, this means that the angle is split into two equal parts or the angle is halved.

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Hence, the $m\angle DBC=25^{\circ}$ .

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3 years ago

The mean annual cost of an automotive insurance policy is normally distributed with a mean of $1140 and standard deviation of $3

DerKrebs [107]

Using the normal distribution, it is found that the probabilities are given as follows:

a) 0.8871 = 88.71%.

b) 0.0778 = 7.78%.

c) 0.8485 = 84.85%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

The parameters in this problem are given as follows:

$\mu = 1140, \sigma = 310, n = 16, s = \frac{310}{\sqrt{16}} = 77.5$

Item a:

The probability is the <u>p-value of Z when X = 1250 subtracted by the p-value of Z when X = 1000</u>, hence:

X = 1250:

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1250 - 1140}{77.5}$

Z = 1.42

Z = 1.42 has a p-value of 0.9222.

X = 1000:

$Z = \frac{X - \mu}{s}$

$Z = \frac{1000 - 1140}{77.5}$

Z = -1.81

Z = -1.81 has a p-value of 0.0351.

0.9222 - 0.0351 = 0.8871 = 88.71% probability.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 1250</u>, hence:

1 - 0.9222 = 0.0778 = 7.78%.

Item c:

The probability is the <u>p-value of Z when X = 1220</u>, hence:

$Z = \frac{X - \mu}{s}$

$Z = \frac{1220 - 1140}{77.5}$

Z = 1.03

Z = 1.03 has a p-value of 0.8485.

0.8485 = 84.85% probability.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

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2 years ago