All categories

3 years ago

Classify each angle Name Each Angle in Four Ways

Mathematics

2 answers:

Sergeu [11.5K]3 years ago

7 0

Well the first one is obtuse and more than 90 degrees, the 2nd one is also obtuse and more than 90 degrees, the 3rd one is acute and less than 90 degrees, the 4th one is obtuse and 180 degrees, and the 5th one is a right angle and is 90 degrees

8_murik_8 [283]3 years ago

5 0

Answer:

1) Obtuse angle

2) Obtuse angle

3) Acute angle

You can do 4 and 5. I will give you a hint that acute angles are less than 90 degrees and obtuse angles are more than 90 degrees. And if it is a straight line it is 180 degrees.

6) Angle KLM, MLK, Obtuse angle, and angle 5.

7) Angle DCB, BCD, Obtuse angle, and angle 5.

You can do 8. I will give you a hint that the letters will start from end-middle-end and it can be opposite. If getting confused then look what I did. On muber 6, my I typed angle KLM and angle MLK. Both of them are same but look on how I typed. Also while naming angles, they will be in capital.

Hope this helps, thank you !!

You might be interested in

A geometry class is asked to find the equation of a line that is parallel to y – 3 = −(x + 1) and passes through (4, 2). Trish s

d1i1m1o1n [39]

Its C

Both students are correct

3 0

4 years ago

Read 2 more answers

I wear my snow boots If and only if it snows.

hoa [83]

Yes u only we're it if itsnows

6 0

4 years ago

Assume that both populations are normally distributed. a) Test whether mu 1 not equals mu 2 at the alpha equals 0.05 level of s

Radda [10]

Answer:

$(16.2-14.1) -2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = -0.445$

$(16.2-14.1) +2.03 \sqrt{\frac{4.5^2}{19} +\frac{3.1^2}{19}} = 4.645$

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".

$n_1 = 19$ sample size 1

$n_2 = 19$ sample size 2

$\\bar X_1 = 16.2$ sample mean for group 1

$\\bar X_2 = 14.2$ sample mean for group 2

$s_1 = 4.5$ sample deviation for group 1

$s_2 = 3.1$ sample deviation for group 2

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

And the degrees of freedom are given by:

$df = n_1 +n_2 -2 = 19+19 -2= 36$

We want a 95% of confidence o then the significance level is 1-0.95 =0.05 and $\alpha/2 = 0.025$ if we find a critical value in the t distribution with 36 degrees of freedom we got: