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slega [8]
1 year ago
10

What is true about AngleQRT? Select two options.

Mathematics
1 answer:
stiv31 [10]1 year ago
5 0

The complete question is

"Triangle QST is isosceles, and Line segment R T bisects AngleT. Triangle Q S T is cut by bisector R T. The lengths of sides S T and Q T are congruent. Line segments S R and R Q are congruent. Angles S T R and R T Q are congruent. What is true about AngleQRT?

Select two options. Measure of angleQRT = 90° Measure of angleQRT = Measure of angleSRT AngleQRT Is-congruent-to AngleSTQ Measure of angleQRT = 2*Measure of angleRTQ AngleQRT Is-congruent-to AngleRTQ"

The true about AngleQRT options are A abd B; ∠QRT = 90 and ∠QRT = ∠SRT

<h3>What is the congruent triangle?</h3>

Two triangles are said to be congruent if the length of the sides is equal, a measure of the angles are equal and they can be superimposed.

Given information;

The triangle QST is isosceles with QT ≅ ST.

The ∠STR and ∠RTQ are congruent.

Now, consider two triangles ∠QTR and ∠STR

In the following triangle

TR  ≅  TR

QTR ≅ STR

QT ≅ ST

Now, according to SAS rule, two congruent triangles have congruent corresponding to their sides,

1. QR ≅ SR

2. QRT ≅ SRT

Since the above two angles are congruent so, they will have the same measure

That will be 90 degree.

Hence, The correct options are; ∠QRT = 90 and ∠QRT = ∠SRT

For more information about congruent visit;

brainly.com/question/13367096

#SPJ1

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You need to use the compound interest formula here.

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The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 mi
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Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions are 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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 51200, \sigma = 8200

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

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

Z = \frac{65000 - 51200}{8200}

Z = 1.68

Z = 1.68 has a pvalue of 0.954

X = 55000

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

Z = \frac{55000 - 51200}{8200}

Z = 0.46

Z = 0.46 has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

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

Z = \frac{48000 - 51200}{8200}

Z = -0.39

Z = -0.39 has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

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

Z = \frac{41000 - 51200}{8200}

Z = -1.24

Z = -1.24 has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

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

Z = \frac{61200 - 51200}{8200}

Z = 1.22

Z = 1.22 has a pvalue of 0.889

X = 41200

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

Z = \frac{41200 - 51200}{8200}

Z = -1.22

Z = -1.22 has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

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2 years ago
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