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faltersainse [42]
4 years ago
7

Which polar equation represents an ellipse?

Mathematics
1 answer:
tatuchka [14]4 years ago
5 0

Answer:

r=\frac{1}{3+2cos\theta}

Step-by-step explanation:

Let us write the equations in standard form:

r=\frac{1}{3+2cos\theta} \implies r=\frac{\frac{1}{3} }{1+\frac{2}{3}\cos \theta }

We have

e=\frac{2}{3}\: and

 ep=\frac{1}{3}

Since the eccentricity  of this conic is less than 1, the conic represents an ellipse.

The second equation is r=\frac{3}{2+3\sin \theta}.

This is a hyperbola, because eccentricity is more than 1.

The third equation is r=\frac{5}{2+2\sin \theta}.

This is a parabola, because eccentricity is  1.

The fourth equation is r=\frac{2}{2-3\sin \theta}.

This is also a hyperbola, because eccentricity is more than 1.

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Mumz [18]

Answer:

Minimum value = 43, Q_1=52, Q_2=Median=61, Q_3=67, Maximum value = 69.

Distribution is negative skewed.

Step-by-step explanation:

The given data set is

44, 43, 67, 68, 57, 65, 52, 62, 69, 54, 53, 61, 67, 48, 65​

Arrange the data in ascending order

43, 44, 48, 52, 53, 54, 57, 61, 62, 65, 65, 67, 67, 68, 69

Divide the data in 4 equal parts.

(43, 44, 48), 52, (53, 54, 57), 61, (62, 65, 65), 67, (67, 68, 69)

Minimum value = 43

Q_1=52

Q_2=Median=61

Q_3=67

Maximum value = 69

Starting and end point of box plot represent the minimum and maximum value respectively. Box starts from the first quartile to the third quartile. A vertical line goes through the box at the median.

The shape of the distribution.

Normally distributed: If Q_3-Q_2=Q_2-Q_1

Positive skew: If Q_3-Q_2>Q_2-Q_1

Negative skew: If Q_3-Q_2

For the given data set

Q_3-Q_2=67-61=6

Q_2-Q_1=61-52=9

Since Q_3-Q_2, therefore the given distribution is negative skewed.

5 0
4 years ago
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(4 - 2, (2 - 1) ----> (2, 1)

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That's the last choice.

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3 years ago
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Answer:

x = 18/5

Step-by-step explanation:

-5/6 x=-3

Multiply each side by -6/5

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Step-by-step explanation:

12 5/6 - 7 1/3

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