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swat32
2 years ago
8

The graph of g(x) is a transformation of the graph of f(x)=3x. Enter the equation for g(x) in the box. G(x) =.

Mathematics
1 answer:
dalvyx [7]2 years ago
8 0

Function transformation involves changing the form of a function

The function g(x) is \mathbf{g(x) = 8(2)^x}

The function is given as:

\mathbf{f(x) = 3^x}

g(x) is an exponential function that passes through points (-2,2) and (-1,4).

An exponential function is represented as:

\mathbf{y = ab^x}

At point (-2,2), we have:

\mathbf{2 = ab^{-2}}

At point (-1,4), we have:

\mathbf{4 = ab^{-1}}

Divide both equations

\mathbf{\frac 42=\frac{ab^{-1}}{ab^{-2}}}

Simplify

\mathbf{2=\frac{b^{-1}}{b^{-2}}}

Apply law of indices

\mathbf{2=b^{-1+2}}

\mathbf{2=b}

Rewrite as:

\mathbf{b =2}

Substitute 2 for b in \mathbf{2 = ab^{-2}}

\mathbf{2 =a(2^{-2})}

This gives

\mathbf{2 =a(\frac 14)}

Multiply both sides by 4

\mathbf{a = 8}

Substitute 8 for (a) and 2 for (b) in \mathbf{y = ab^x}

\mathbf{y = 8(2)^x}

Express as a function

\mathbf{g(x) = 8(2)^x}

Hence, the function g(x) is \mathbf{g(x) = 8(2)^x}

Read more about exponential functions at:

brainly.com/question/11487261

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Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of
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Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

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p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

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Answer:

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Step-by-step explanation:

(X+4)(x+2)

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4+2=6

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Step-by-step explanation:

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