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3 years ago

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1 answer:

Lorico [155]3 years ago

6 0

Answer:

OK I hear you well

Thanks for you

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Can someone please help its due soon.ill give brainliest!

sammy [17]

Answer: The Second one

Step-by-step explanation:

The second answer selection tells us that angle VXQ and ZXY are congruent. That's one angle we know that is congruent.

The proportion of YX/WX and WX/YX is the same

(16/12 = 4/3 and 12/9 = 4/3)

So now we know that two sides are congruent to each other. This is enough to satisfy the side side angle congruency test.

4 0

3 years ago

Which of the following is NOT a variation of a Pythagorean identity?

Ipatiy [6.2K]

The expression that is not a variation of the Pythagorean identity is the third option.

<h3>What is the Pythagorean identity?</h3>

The Pythagorean identity can be written as:

$sin^2(x) + cos^2(x) = 1$

For example, if we subtract cos^2(x) on both sides we get the second option:

$sin^2(x) = 1 - cos^2(x)$

Which is a variation.

Now, let's divide both sides by cos^2(x).

$sin^2(x)/cos^2(x) + cos^2(x)/cos^2(x) = 1/cos^2(x)\\\\tan^2(x) + 1 = sec^2(x)\\\\tan^2(x) - sec^2(x) = -1$

Notice that the third expression in the options looks like this one, but the one on the right side is positive. The above expression is in did a variation of the Pythagorean identity, then the one written in the options (with the 1 instead of the -1) is incorrect, meaning that it is not a variation of the Pythagorean identity.

Concluding, the correct option is the third one.

If you want to learn more about the Pythagorean identity, you can read:

brainly.com/question/24287773

3 0

3 years ago

Is this rightttttttttttttttt please tell me

Snowcat [4.5K]

Answer:

Yes!

Step-by-step explanation:

6 0

3 years ago

In triangle PQR, Line QS and line SR are angle bisectors to angle P (74 degrees). How many Degrees are in the angle QSR.

AlexFokin [52]

we know that

<u>in the triangle PQR</u>

m∠PQR+m∠QPR+m∠QRP=180°------> the sum of the internal angles of a triangle is equal to 180 degrees

(m∠PQR+m∠QRP)+74°=180°-------> (m∠PQR+m∠QRP)=106°

we know that

If QS and SR are angle bisectors

then

<u>In the triangle QSR</u>

m∠SQR+m∠QSR+m∠QRS=180°-------> equation 1

and

(m∠SQR+m∠QRS)=106°/2--------> (m∠SQR+m∠QRS)=53°

substitute the value of (m∠SQR+m∠QRS)=53° in the equation 1

53°+m∠QSR=180°

m∠QSR=180°-53°-------->m∠QSR= 27°

therefore

<u>the answer is</u>

The measure of angle QSR is 27 degrees

4 0

4 years ago

The weights of 83 randomly selected windshields were found to have a variance of 1.88. Construct the 95% confidence interval for

Morgarella [4.7K]

Answer:

95% confidence interval for the population variance = (1.42 , 2.62).

Step-by-step explanation:

We are given that the weights of 83 randomly selected windshields were found to have a variance of 1.88.

<em>So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;</em>

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, $s^{2}$ = sample variance = 1.88

$\sigma^{2}$ = population variance

n = sample of windshields = 83

So, 95% confidence interval for population variance, $\sigma^{2}$ is;

P(58.85 < $\chi^{2} __8_2$ < 108.9) = 0.95 {As the table of $\chi^{2}$ at 82 degree of freedom

gives critical values of 58.85 & 108.9}

P(58.85 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 108.9) = 0.95

P( $\frac{ 58.85}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{108.9}{(n-1)s^{2} }$ ) = 0.95

P( $\frac{ (n-1)s^{2}}{108.9 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{58.85 }$ ) = 0.95

<em><u>95% confidence interval for</u></em> $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{108.9 }$ , $\frac{ (n-1)s^{2}}{58.85 }$ )

= ( $\frac{ (83-1)\times 1.88}{108.9 }$ , $\frac{ (83-1)\times 1.88}{58.85 }$ )

= (1.42 , 2.62)

Therefore, 95% confidence interval for the population variance of the weights of all windshields in this factory is (1.42 , 2.62).

8 0

3 years ago