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2 years ago

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Mathematics

2 answers:

fredd [130]2 years ago

6 0

Answer:

Median: 27

Mean: 27

Step-by-step explanation:

ArbitrLikvidat [17]2 years ago

5 0

Answer:

For both sets of data, the median is equal to the mean.

The interquartile range (IQR) is greater for M's classes than J's classes.

Step-by-step explanation:

<u>J's Classes:</u>

23, 25, 27, 28, 32

Median: 27

Mean: 27

Mean Absolute Deviation: 2.4

Interquartile Range: 3

<u>M's Classes:</u>

20, 24, 26, 29, 31

Median: 26

Mean: 26

Mean Absolute Deviation: 3.2

Interquartile Range: 5

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ArbitrLikvidat [17]

Answer:

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , c) $x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$ , $y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$ .

Step-by-step explanation:

The equation of the circle is:

$x^{2} + (y-1)^{2} = 16$

After some algebraic and trigonometric handling:

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1$

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t$

Where:

$\frac{x}{4} = \cos t$

$\frac{y-1}{4} = \sin t$

Finally,

$x = 4\cdot \cos t$

$y = 1 + 4\cdot \sin t$

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

c) $x = 4\cdot \cos t''$ , $y = 1 + 4\cdot \sin t''$

Where:

$4\cdot \cos t' = 0$

$1 + 4\cdot \sin t' = 5$

The solution is $t' = \frac{\pi}{2}$

The parametric equations are:

$x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$

$y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$

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