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Fantom [35]
2 years ago
10

Quadrilateral OPQR is inscribed in circle N, as shown below. Which of the following could be used to calculate the measure of ∠P

QR?
Circle N is shown with a quadrilateral OPQR inscribed inside it. Angle O is labeled x plus 16. Angle P is not labeled. Angle Q is labeled 6x minus 4. Angle R is labeled 2x plus 16.

(6x − 4)° + (2x + 16)° = 180°
(x + 16)° + (6x − 4)° = 180°
(6x − 4)° + (2x + 16)° = 360°
(x + 16)° + (6x − 4)° = 360°
Mathematics
2 answers:
Anna [14]2 years ago
8 0

The equation to be used to calculate the measure of ∠PQR in the cyclic quadrilateral is: (6x − 4)° + (2x + 16)° = 180°.

<h3>What are Opposite Angles in a Cyclic Quadrilateral?</h3>

A cylcic quadrilateral is a quadrilateral inscribed in a circle. The opposite angles are supplementary, that is they add up to give 180 degrees.

Thus, angles Q and O are opposite angles in the cyclic quadrilateral, which means they are supplementary.

Therefore, the equation to be used to calculate the measure of ∠PQR in the cyclic quadrilateral is: (6x − 4)° + (2x + 16)° = 180°.

Learn more about cyclic quadrilateral on:

brainly.com/question/10057464

sammy [17]2 years ago
5 0

Answer:

I just took the quiz and the answer above is NOT correct.

Step-by-step explanation:

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A violin student records the number of hours she spends practicing during each of nine consecutive weeks: 6.2 5.0 4.3 7.4 5.8 7.
spin [16.1K]

Answer:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first  quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

Step-by-step explanation:

Assuming this complete question: A violin student records the number of hours she spends practicing during each of nine consecutive weeks:

6.2 5.0 4.3 7.4 5.8 7.2 8.4 1.2 6.3

For this case we need to sort the data first on increasing way and we got:

1.2, 4.3, 5.0, 5.8, 6.2, 6.3, 7.2, 7.4, 8.4

For this case we have 9 values the median would be on the 5 position:

Median = 6.2

The first quartile would be 5 since we analyze 1.2, 4.3, 5.0, 5.8, 6.2 and the middle point is 5.0

The third quartile would be 7.2 since we analyze 6.2, 6.3, 7.2, 7.4, 8.4 and the middle point is 7.2

Then the interquartile rnage would be:

IQR = Q_3 -Q_1= 7.2-5=2.2

And 1.5 IQR = 1.5*2.2=3.3

The lower limit on this case would be:

LL= Q_1 -1.5 IQR= 5-3.3=1.7

And our value is 1.2 is lower than the lower limit 1.7

Considering the smallest data value (1.2) and using the 1.5 × IQR rule, we would:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first  quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

B) not classify the value 1.2 as an outlier, because it is not more than 1.5 × IQR below  the first quartile.

False 1.2 is more than 1.5 IQR below the first quartile

C) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the  median.

False, we analyze if is an outlier comparing with the Q1 or Q3 not with the median

D) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the  mean.

False we analyze if is an outlier comparing with the Q1 or Q3 not with the mean

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