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astraxan [27]
3 years ago
14

What transformations change the graph of f(x) to the graph of g(x)?

Mathematics
2 answers:
Dennis_Churaev [7]3 years ago
3 0

For this case we have that the main function is given by:

f (x) = - 7x ^ 2

We apply the following transformations:

Vertical expansions:

To graph y = a * f (x)

If a> 1, the graph of y = f (x) is expanded vertically by a factor a.

For a = 5 we have:

h (x) = 5 * f (x)\\h (x) = 5 * (- 7x ^ 2)\\h (x) = - 35x ^ 2

Vertical translations:

Suppose that k> 0

To graph y = f (x) + k, move the graph of k units up.

For k = 5 we have:

g (x) = h (x) + k\\g (x) = - 35x ^ 2 + 5

Answer:

The graph of g (x) is the graph of f (x) stretched vertically by a factor of 5 and translated up 5 units.

joja [24]3 years ago
3 0
I think the correct answer from the choices listed above is the last option. The transformation change of the graph of f(x) to the graph of g(x) would be that the graph of g(x) is the graph of f(x) stretched vertically by a factor of 5 and translated up 5 units.
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In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p
skelet666 [1.2K]

Answer:

\mathbf{P(X=5) =0.0888}    

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

\mathtt{P(X=x) =(^{n}_{x} )   \  \pi^x \  (1-\pi)^{n-x}}

\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} )   \  \pi^x \  (1-\pi)^{n-x}}

where;

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The probability \mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} )   \  0.36^5 \  (1-0.36)^{8-5}}

\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} )   \  0.36^5 \  (0.64)^{3}}

\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =({8 \times 7 } )  \times  \ 0.0060466 \  \times 0.262144}

\mathtt{P(X=5) =0.0887645}

\mathbf{P(X=5) =0.0888}     to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5)\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})

{P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times  (0.36)^0  \times  (1-0.36)^8  \ )  +  \dfrac{8!}{1!(7!)} \times  (0.36)^1  \times  (1-0.36)^7  \ +\dfrac{8!}{2!(6!)} \times  (0.36)^2  \times  (1-0.36)^6  \ +  \dfrac{8!}{3!(5!)} \times  (0.36)^3  \times  (1-0.36)^5 +  \dfrac{8!}{4!(4!)} \times  (0.36)^4  \times  (1-0.36)^4  \  +  \dfrac{8!}{5!(3!)} \times  (0.36)^5  \times  (1-0.36)^3  \ )

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1  - P( x  ≤ 5 )

P ( x ≥ 6) = 1  - 0.9707

P ( x ≥ 6) = 0.0293

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