Notice that, the angle QRS is external to the triangle and adjacent to the angle PRQ. According to the theorem of a external/adjacent angle, we have: m∠QRS = m∠PQR + m∠RPQ, where PQR and RPQ are internal angles.

From the hypothesis, we have:

m∠QRS =(10x−12)∘(10x−12)

m∠PQR = (3x+20)∘(3x+20)

m∠RPQ=(3x−8)∘(3x−8)

Using the first equation and replacing the hypothesis:

m∠QRS = m∠PQR + m∠RPQ

(10x−12)∘(10x−12) = (3x+20)∘(3x+20) + (3x−8)∘(3x−8)

Multiplying and applying the remarkable identity:

Then, we use a calculator to find the roots, which are:

In this case, we will see what root is the right one.

Now, we replace it into m∠QRS =(10x−12)∘(10x−12), because we need to find m∠QRS.

m∠QRS =(10x−12)∘(10x−12) = (10(4.7) - 12) (10(4.7) - 12) = (35) (35) = 1225

Step-by-step explanation:

8 0

3 years ago

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A consumer group surveyed 146 airplane travelers after a flight and found that 132 of them would fly that airline again. Find th

prohojiy [21]

Answer:

$\hat p =\frac{X}{n}$

And replacing we got:

$\hat p =\frac{132}{146}= 0.904$

And for this case the standard error assuming normality would be given by:

$SE= \sqrt{\frac{\hat p (1-\hat p)}{n}}$

And replacing we got:

$SE= \sqrt{\frac{0.904*(1-0.904)}{146}}= 0.024$

Step-by-step explanation:

For this problem we know the following notation:

$n= 146$ represent the sample size selected

$X= 132$ represent the number of airplane travelers who after a flight would fly that airline again

The estimated proportion for this case would be:

$\hat p =\frac{X}{n}$

And replacing we got:

$\hat p =\frac{132}{146}= 0.904$

And for this case the standard error assuming normality would be given by:

$SE= \sqrt{\frac{\hat p (1-\hat p)}{n}}$

And replacing we got:

$SE= \sqrt{\frac{0.904*(1-0.904)}{146}}= 0.024$

5 0

3 years ago