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VashaNatasha [74]
3 years ago
11

An equilateral triangle and an isosceles triangle share a common side. What is the measure of angle ABC?

Mathematics
1 answer:
o-na [289]3 years ago
4 0

Answer:

Therefore ,the measure of angle ABC = 116°.

Step-by-step explanation:

Given:

Δ CBD is an Equilateral Triangle

BC = BD = CD

Δ ABD is an Isosceles Triangle

AB = BD

To Find:

m∠ ABC = ?

Solution:

Property For an Equilateral Triangle:

All the measure of Each of the angles angles of a triangle is 60°.

Δ CBD is an Equilateral Triangle   ..........Given

∴ m∠ CBD = 60° ....................( 1 )

Property For an Isosceles Triangle:

Any two of the base angles are equal.

Δ ABD is an Isosceles Triangle ................Given

∴ m∠ BAD = m∠ BDA = 62°  .....................( 2 )

Property of Triangle:

SUM of the measure of an angles of a triangle is 180°

In Δ BAD,

m∠ ABD + m∠ BAD + m∠ BDA = 180°

Substituting the values from equation ( 2 ) in it we get

m∠ ABD + 62 + 62 = 180

∴ m∠ ABD = 180 -124

∴ m∠ ABD = 56°     ................( 3 )

Now By angle addition property we have

m∠ ABC = m∠ ABD  + m∠ CBD

Substituting the values from equation 1 and equation 3 we get

∴ m∠ ABC = 56 + 60

∴ m∠ ABC = 116°

Therefore ,the measure of angle ABC = 116°.

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Answer:

It does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

Step-by-step explanation:

We are given the following data in the question:

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Formula:

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where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{1067}{15} = 71.13

Sum of squares of differences = 4739.733

S.D = \sqrt{\frac{4739.733}{14}} = 18.39

Population mean, μ = 60 minutes

Sample mean, \bar{x} = 71.13 minutes

Sample size, n = 15

Alpha, α = 0.10

Sample standard deviation, s = 18.39 minutes

First, we design the null and the alternate hypothesis

H_{0}: \mu = 60\text{ minutes}\\H_A: \mu \neq 60\text{ minutes}

We use Two-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{71.13 - 60}{\frac{18.39}{\sqrt{15}} } = 2.34

Calculating the p-value from the table, we have,

P-value = 0.034354

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, we conclude that it does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

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Step-by-step explanation:

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