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sleet_krkn [62]
3 years ago
15

For the inverse variation function y = k/x (where x, k > 0), what happens to the value of y as the value of x increases?

Mathematics
2 answers:
Allisa [31]3 years ago
7 0
As the value of x increases, y decreases.

ser-zykov [4K]3 years ago
4 0

Answer:

the answer is B

Step-by-step explanation:


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Six less than seven times a number is equal to -20
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Let this number be x.
"Six less than seven times a number is equal to -20" can be represented by the equation 7x - 6 = -20. We need to solve for x.
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4 0
4 years ago
In a certain city, the daily consumption of electric power, in millions of kilowatt-hours, is a random
kupik [55]

Answer:

a) \alpha = 3, \beta = 2

b) 0.0620

Step-by-step explanation:

We are given the following in the question:

Population mean, \mu = 6

Variance, \sigma^2 = 12

a) Value of \alpha, \beta

We know that

\alpha \beta = \mu = 6\\\alpha \beta^2 = \sigma^2 = 12

Dividing the two equations, we get,

\dfrac{\alpha\beta^2}{\alpha\beta} = \dfrac{12}{6}\\\\\Rightarrow \beta = 2\\\alpha \beta = 6\\\Rightarrow \alpha = 3

b) probability that on any given day the daily  power consumption will exceed 12 million kilowatt hours.

We can write the probability density function as:

f(x,3,2) = \dfrac{1}{2^{3}(3-1)!}x^{3-1}e^{-\frac{x}{2}}, x > 0\\\\f(x,3,2) = \dfrac{1}{16}x^{2}e^{-\frac{x}{2}}, x > 0

We have to evaluate:

P(x >12)\\\\= \dfrac{1}{16}\displaystyle\int^{\infty}_{12}f(x)dx\\\\=\dfrac{1}{16}\bigg[-2x^2e^{-\frac{x}{2}}-2\displaystyle\int xe^{-\frac{x}{2}}dx}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[x^2e^{-\frac{x}{2}}+4xe^{-\frac{x}{2}}+8e^{-\frac{x}{2}}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[(\infty)^2e^{-\frac{\infty}{2}}+4(\infty)e^{-\frac{\infty}{2}}+8e^{-\frac{\infty}{2}} -( (12)^2e^{-\frac{12}{2}}+4(12)e^{-\frac{12}{2}}+8e^{-\frac{12}{2}})\bigg]\\\\=0.0620

0.0620 is the required probability that on any given day the daily  power consumption will exceed 12 million kilowatt hours.

3 0
4 years ago
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