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Pani-rosa [81]
1 year ago
5

function g is a transformation of the parent exponential function f(x)=x^2. the graph of g is a translation right 5 and up 1 uni

t of the graph of f. write the equation for g in thr form y=ax^2+bx+c
Mathematics
1 answer:
adell [148]1 year ago
4 0

The equation of g in the form y = ax² + bx +c is g(x) =  x² - 10x + 26

What is a function ?

A function is a kind of rule that, for one input, it gives you one output. Image source: by Alex Federspiel. An example of this would be y=x2. If you put in anything for x, you get one output for y. We would say that y is a function of x since x is the input value.

Horizontal translation:

For a general function f(x) an horizontal translation of N units is written as:

g(x) = f(x + N)

If N is positive, the shift is to the left

If N is negative, the shift is to the right.

Vertical translation:

For a general function f(x) a vertical translation of N units is written as:

g(x) = f(x) + N

If N is positive, the shift is upwards

If N is negative, the shift is downwards.

First, we apply a translation to the right of 5 units, then at the moment we have:

g(x) = f(x - 5)

Then we translate the graph up 1 units, then:

g(x) = f(x - 5) + 1

Now we can replace it by the actual function to get:

g(x) = (x - 5)^2 + 1

      =  x² - 2*5*x + 25 + 1

      = x² - 10x + 26

To learn more about a function from the given link

brainly.com/question/22340031

#SPJ9  

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A statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20. Use thi
rjkz [21]

Answer:

\mu = 14.5\\

\sigma = 5.071\\

k = 1.084

Step-by-step explanation:

given that a  statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20.

i.e. his findings with respect to probability are

P(9

Recall Chebyshev's inequality that

P(|X-\mu |\geq k\sigma )\leq {\frac {1}{k^{2}}}\\P(|X-\mu |\leq k\sigma )\geq 1-{\frac {1}{k^{2}}}\\

Comparing with the Ii equation which is appropriate here we find that

\mu =14.5

Next what we find is

k\sigma = 5.5\\1-\frac{1}{k^2} =0.15\\\frac{1}{k^2}=0.85\\k=1.084\\1.084 (\sigma) = 5.5\\\sigma = 5.071

Thus from the given information we find that

\mu = 14.5\\\sigma = 5.071\\k = 1.084

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