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3 years ago

Which of the following can be used to prove that triangle EFG is also a right triangle?

Mathematics

2 answers:

galben [10]3 years ago

4 0

Answer:

Prove that KL is equal to c by Pythagorean Theorem.

Step-by-step explanation:

In the two triangles triangle KLM is a right triangle with sides a and b as the two legs of the right triangle .We can apply Pythagorean Theorem to find KL.

In triangle KGF two sides are a and b .If we prove KL=EF then Pythagorean Theorem will be true for triangle KGF and hence we can prove triangle KGF to be a right triangle.

The correct option to prove triangle KGF to be a right triangle is to Prove that KL is equal to c by Pythagorean Theorem.

Vika [28.1K]3 years ago

4 0

Which of the following can be used to prove that triangle EFG is also a right triangle?

Prove that the sum of a and c is greater than b.
Prove that the sum of a and b is greater than c.
Prove that the ratio of EF and KL is greater than 1 and hence, the triangles are similar by AA postulate.
Prove that KL is equal to c by Pythagorean Theorem.

The correct answer is:
Prove that KL is equal to c by Pythagorean Theorem.

It can also be "Prove that triangles are congruent by SSS property and hence angle EGF is equal to angle KML"

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Please solve the whole page please please

Leya [2.2K]

1d) ↑ (Do the same but turned if that's easier)

1a) 199

1b) 193

1c) 217

1e) Based on the box plot, we can for example, find the lowest and greatest values, the range (largest - smallest), the mean (all values / # values), and the median middle value or q2 given by the second line in the box.

2b) 7

2c) 9.5

2d) ↑

2e) because the data is not ordered, it is difficult to determine the quartile data. Once it is ordered like 6, 7, 7, 7, 8, 8, 9, 10, 10. The data can be visualized alot easier. With the box plot we can determine q3 of the data by looking at the 3rd line, using the whiskers of the box to find the minimum and maximum values.

You need to find the minimum, maximum, the quartile data obtained by having an even amount of data on both sides, order it so that values can be grouped. this includes q1, q2 (or median), and q3.

In this case the ordered data would be: 4,6,8,8,9,11,12,14,14,16.

[4, 6, 8, 8, 9] [] [11, 12, 14, 14, 16]

4 is the minimum, 16 is the maximum, q1 is 8, q3 is 14, and median is 10.

4) ↑

You need to find the minimum, maximum, the quartile data obtained by having an even amount of data on both sides, order it so that values can be grouped. this includes q1, q2 (or median), and q3.

In this case the ordered data would be:

45, 45, 50, 55, 60, 70, 70, 75, 80, 85

[45, 45, 50, 55, 60] [] [70, 70, 75, 80, 85]

45 is the minimum, 85 is the maximum, q1 is 50, q3 is 75, and median is 65.

6) ↑

4 0

2 years ago

Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - α)% confidence interval for the population variance is given as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

It is provided that:

n = 20

s = 3.9

Confidence level = 95%

⇒ α = 0.05

Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$