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Softa [21]
2 years ago
7

What is the equation of a hyperbola with a = 1 and c = 9? Assume that the transverse axis is horizontal.

Mathematics
1 answer:
belka [17]2 years ago
8 0

Answer:

Step-by-step explanation:

The general formula for a horizontal hyperbola :

\frac{x^{2} }{a^{2} } -\frac{y^{2} }{b^{2} } =1

Given: a = 1 we know a² = 1 ,and that c = 9 so we know c² = 9² = 81

We also know that the relation between a, b, c for a hyperbola is c²= a²+b²

c²= a²+b², substitute what we know

81 = 1 +b², subtract 1 from both sides of the equation

80 = b²

The equation of our hyperbola is:

\frac{x^{2} }{1 } -\frac{y^{2} }{80 } =1  or  x^{2} -\frac{y^{2} }{80} =1

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explanation-
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Answer:

<h2>Second table</h2>

Step-by-step explanation:

<em>One-to-one function is a function that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain. Every element of the function's domain is the image of at most one element of its domain.</em>

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7 0
3 years ago
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One measure of an athlete’s ability is the height of his or her vertical leap. Many professional basketball players are known fo
almond37 [142]

Answer:

(1) P(\bar X < 26 inches) = 0.0436

(2) P(27.5 inches < \bar X < 28.5 inches) = 0.2812

Step-by-step explanation:

We are given that the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.

Firstly, Let \bar X = mean vertical leap for the 36 players

Assuming the data follows normal distribution; so the z score probability distribution for sample mean is given by;

            Z = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean vertical  leap = 28 inches

            \sigma = standard deviation = 7 inches

            n = sample of NBA player = 36

(1) Probability that the mean vertical leap for the 36 players will be less than 26 inches is given by = P(\bar X < 26 inches)

   P(\bar X < 26) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{26-28}{\frac{7}{\sqrt{36} } } ) = P(Z < -1.71) = 1 - P(Z \leq 1.71)

                                                 = 1 - 0.95637 = 0.0436

(2) <em>Now, here sample of NBA players is 26 so n = 26.</em>

Probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is given by = P(27.5 inches < \bar X < 28.5 inches) = P(\bar X < 28.5 inches) - P(\bar X \leq 27.5 inches)

    P(\bar X < 28.5) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{28.5-28}{\frac{7}{\sqrt{26} } } ) = P(Z < 0.36) = 0.64058 {using z table}                      

    P(\bar X \leq 27.5) = P( \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{27.5-28}{\frac{7}{\sqrt{26} } } ) = P(Z \leq -0.36) = 1 - P(Z < 0.36)

                                                        = 1 - 0.64058 = 0.35942

Therefore, P(27.5 inches < \bar X < 28.5 inches) = 0.64058 - 0.35942 = 0.2812

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