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3 years ago

In the figure below line DE is parallel to line FG and transversal AG

Mathematics

1 answer:

Arturiano [62]3 years ago

4 0

Answer: The answer is 147° and 33°.

Step-by-step explanation: We are given a figure in which the line DE is parallel to the line FG and AG is a tranversal. We are to find the value of 'x' and 'y' from the figure.

Also, given that

∠ABE = (5y - 18)°.

We can see from the figure that ∠DBG and ∠ABE are vertically opposite angles, so they must be equal.

That is,

$\angle DBG=\angle ABE\\\\\Rightarrow x=5y-18\\\\\Rightarrow 5y-x=18.~~~~~~~~~~~~~~~(i)$

Also, since ∠DBG and ∠FGB are interior angles on the same side of the transversal, so their sum is 180°.

That is

$\angle DBG+\angle FGB=180^\circ\\\\\Rightarrow x+y=180.~~~~~~~~~~~~~~~~~(ii)$

Adding equations (i) and (ii), we get

$y+5y=18+180\\\\\Rightarrow 6y=198\\\\\Rightarrow y=33,$

and from equation (ii), we get

$x=180-33=147.$

Thus, x° = 147° and y° = 33°.

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Ten college students were randomly selected. Their grade point averages (GPAs) when they entered the program were between 3.5 a

solong [7]

Answer:

The 95% confidence interval for the true slope is (0.03985, 0.14206).

Step-by-step explanation:

For the regression equation:

$\hat y=\alpha +\hat \beta x$

The (1 - <em>α</em>)% confidence interval for the regression coefficient or slope $(\hat \beta )$ is:

$Ci=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )$

The regression equation for current GPA (Y) of students based on their GPA's when entering the program (X) is:

$\hat Y=3.584756+0.090953 X$

The summary of the regression analysis is:

Predictor Coefficient SE t-stat p-value

Constant 3.584756 0.078183 45.85075 5.66 x 10⁻¹¹

Entering GPA 0.090953 0.022162 4.103932 0.003419

The regression coefficient and standard error are:

$\hat \beta = 0.090953\\SE (\hat \beta)=0.022162$

The critical value of <em>t</em> for 95% confidence level and 8 degrees of freedom is:

$t_{\alpha/2, n-2}=t_{0.05/2, 10-2}=t_{0.025, 8}=2.306$

Compute the 95% confidence interval for $(\hat \beta )$ as follows:

$CI=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )\\=0.090953\pm 2.306\times 0.022162\\=0.090953\pm 0.051105572\\=(0.039847428, 0.142058572)\\\approx (0.03985, 0.14206)$

Thus, the 95% confidence interval for the true slope is (0.03985, 0.14206).

6 0

3 years ago

please help me, a point thief gave a me a silly fake answer the first time, now no one will answer me. Write the given equation

vaieri [72.5K]

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4 years ago

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Sholpan [36]

2/3.
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3 years ago

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Given the formula PV = nRT , solve for the variable T

aleksley [76]

T=PV/Rn you just divide both sides by Rn

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3 years ago

Write a system of equations to describe the situation below, solve using any method, and fill in the blanks. The volleyball team

almond37 [142]

Answer: After washing 20 cars together, each team will have raised the same amount in total.

Step-by-step explanation:

Let x represent the number of cars that each each teams will wash for them to raise the same amount in total.

The volleyball team gets $4 per car. In addition, they have already brought in $24 from past fundraisers. This means that the total amount raised by the volleyball team after washing x cars would be

4x + 24

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For both amounts to be equal, the number of cars would be

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x = 60/3

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3 years ago