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Stella [2.4K]
3 years ago
5

The average January surface water temperatures (°C) of Lake Michigan from 2000 to 2009 were 5.07, 3.57, 5.32, 3.19, 3.49, 4.25,

4.76, 5.19, 3.94, and 4.34.
The mean value of these temperatures is 4.312.


What is the variance of this data set?
Will give Brainliest

Mathematics
2 answers:
Varvara68 [4.7K]3 years ago
8 0

Answer:

UwU

Step-by-step explanation:

Mnenie [13.5K]3 years ago
6 0

Answer:

0.5192

Step-by-step explanation:

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The center of a hyperbola is located at the origin. One focus is located at (−50, 0) and its associated directrix is represented
leva [86]

The equation of the hyperbola is : \frac{x^{2}}{48^2}  - \frac{y^{2}}{14^2}  = 1

The center of a hyperbola is located at the origin that means at (0, 0) and one of the focus is at (-50, 0)

As both center and the focus are lying on the x-axis, so the hyperbola is a horizontal hyperbola and the standard equation of horizontal hyperbola when center is at origin: \frac{x^{2}}{a^{2}}  - \frac{y^{2}}{b^{2}}    = 1

The distance from center to focus is 'c' and here focus is at (-50,0)

So, c= 50

Now if the distance from center to the directrix line is 'd', then

d= \frac{a^{2}}{c}

Here the directrix line is given as : x= 2304/50

Thus, \frac{a^{2}}{c}  = \frac{2304}{50}

⇒ \frac{a^{2}}{50}  = \frac{2304}{50}

⇒ a² = 2304

⇒ a = √2304 = 48

For hyperbola, b² = c² - a²

⇒ b² = 50² - 48² (By plugging c=50 and a = 48)

⇒ b² = 2500 - 2304

⇒ b² = 196

⇒ b = √196 = 14

So, the equation of the hyperbola is : \frac{x^{2}}{48^2}  - \frac{y^{2}}{14^2}  = 1

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I got -34/5
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4y -7 = 5 -2y +2y +2y add 2y to each side 6y -7 = 5 +7 +7 add 7 to each side y =
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4y - 7 = 5 - 2y

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