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3 years ago

M∥n, m∠1 = 50°, m∠2 = 48°, and line s bisects ∠ABC. What is m∠3?

Mathematics

2 answers:

Nutka1998 [239]3 years ago

5 0

Answer

m<3 = 49°

Step-by-step explanation:

It is given that,

M∥n, m∠1 = 50°, m∠2 = 48°, and line s bisects ∠ABC

m<DEF = m<1 + m<2 = 50° + 48° = 98°

Corresponding angle of <DEF is equal to <ABC = 98°

To find m<3

< 4 = <5 = 98/2 = 49° (Since line s bisects ∠ABC)

Therefore,

m<3 = 49° (< 4 and <3 are vertically opposite angles)

Sindrei [870]3 years ago

5 0

<u>Answer:</u>

m∠3 = 49°

<u>Step-by-step explanation:</u>

We are given that the angles m∠1 = 50° and m∠2 = 48° and the line s bisects ∠ABC.

If m∠1 = 50° and m∠2 = 48°, then ∠DEF = 50 + 48 = 98°

So ∠DEB will be equal to = 180 - 98 = 82°

If ∠DEB = 82°, then the angle from A to B will 180 - 82 = 98°.

We know that the line s bisects ∠ABC, therefore the measure of angle m∠3 will be half of 98 = 49°

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$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$

And replacing we got:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

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3 years ago