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3 years ago

The measure of angle A is 5 times the measure of its supplement.

1 answer:

8 0

Let the angle a be 5x
The supplement be x
According to the question
X+5x=180
6x=180
X=30
5x=5×30=150
The measure of angle a 150
Angle of supplement 30
Hope this helps!!

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Pretty Pavers company is installing a driveway. Below is a diagram of the driveway they are

prohojiy [21]

Answer:

The most correct option is;

(B) 958.2 ft.²

Step-by-step explanation:

From the question, the dimension of each square = 3 ft.²

Therefore, the length of the sides of the square = √3 ft.

Based on the above dimensions, the dimension of the small semicircle is found by counting the number of square sides ti subtends as follows;

The dimension of the diameter of the small semicircle = 10·√3

Radius of the small semicircle = Diameter/2 = 10·√3/2 = 5·√3

Area of the small semicircle = (π·r²)/2 = (π×(5·√3)²)/2 = 117.81 ft.²

Similarly;

The dimension of the diameter of the large semicircle = 10·√3 + 2 × 6 × √3

∴ The dimension of the diameter of the large semicircle = 22·√3

Radius of the large semicircle = Diameter/2 = 22·√3/2 = 11·√3

Area of the large semicircle = (π·r²)/2 = (π×(11·√3)²)/2 = 570.2 ft.²

Area of rectangle = 11·√3 × 17·√3 = 561

Area, A of large semicircle cutting into the rectangle is found as follows;

$A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (\theta - sin\theta) \times r^2$

Where:

$\theta = 2\times tan^{-1}( \frac{The \, number \, of \, vertical \, squrare \, sides \ cut \, by \ the \ large \, semicircle}{The \, number \, of \, horizontal \, squrare \, sides \ cut \, by \ the \ large \, semicircle} )$

$\therefore \theta = 2\times tan^{-1}( \frac{10\cdot \sqrt{3} }{5\cdot \sqrt{3}} ) = 2.214$

Hence;

$A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (2.214 - sin2.214) \times (11\cdot\sqrt{3} )^2 = 128.3 \, ft^2$

Therefore; t

The area covered by the pavers = 561 - 128.3 + 570.2 - 117.81 = 885.19 ft²

Therefor, the most correct option is (B) 958.2 ft.².

4 0

3 years ago

HELP!!!!!!!!!!!!!!!!

koban [17]

275=25x+100 is the correct equation for the number of months it will take for wade to save $275

5 0

3 years ago

Find the geometric sum: 40 + 20 + 10 + … + 0.078125

sveticcg [70]

$S=40+20+10+\cdots+0.078125$
$S=40\left(1+\dfrac12+\dfrac14+\cdots+\dfrac1{512}\right)$
$S=40\left(1+2^{-1}+2^{-2}+\cdots+2^{-9}\right)$

$2^{-1}S=40\left(2^{-1}+2^{-2}+2^{-3}+\cdots+2^{-10}\right)$

$\implies S-2^{-1}S=40\left(1-2^{-10}\right)$
$\implies\dfrac12S=40\left(1-\dfrac1{1024}\right)$
$\implies S=80\left(1-\dfrac1{1024}\right)=\dfrac{5115}{64}=79.921875$

4 0

3 years ago

26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics

hjlf

The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

$\mu _d = \mu _{post} - \mu _{pre}$

Then the null hypotheses and alternative hypotheses will be

H₀: $\mu _d = 0$

Hₐ: $\mu _d > 0$

Then the test statistic will be

$\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5$

Then

$\rm S_d = 6.02$

The test statistic value is given by

$\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16$

Since this is a right-tailed test, so the critical value is given by

$\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353$

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that $\mu _d > 0$ that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

#SPJ1

7 0

2 years ago

Answers needed help me

frozen [14]

Answer:

it's blurry, you need a new picture

5 0

3 years ago