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3 years ago

Which exponential function has an initial value of 2?

Mathematics

2 answers:

Leviafan [203]3 years ago

8 0

Answer:

None of the exponential equations given in the two options have initial value 2.

Step-by-step explanation:

The general form of an exponential function to have initial value of 2 is

$f(x) = 2(b)^{x}$ where b ≠ 1.

Now, in the given two options the first one has equation $f(x) = 3(2)^{x}$ , and the other given table forms the exponential function equation $f(x) = \frac{1}{2}(2)^{x}$ .

Therefore, none of the exponential equations given in the two options have initial value 2. (Answer)

lina2011 [118]3 years ago

7 0

Answer: f(x) = 2(3x)

Step-by-step explanation:

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4 years ago

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larisa86 [58]

Answer:

The 99% confidence interval for the population mean is 22.96 to 26.64

Step-by-step explanation:

Consider the provided information,

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Thus, 1-α=0.99

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Now by using z score table we find that $z_{\frac{\alpha}{2}}=2.58$

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4 years ago

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3 years ago

Find the answer to zero point 73 repeating times square root of 147 end square root and select the correct answer below. Also, i

liubo4ka [24]

Answer:

$\frac{511\,\,\sqrt{3} }{99}$ which is an irrational number

Step-by-step explanation:

Recall that the repeating decimal 0.7373737373... can be written in fraction form as: $\frac{73}{99}$

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$147=7^2\,3$

Then, 7 will be able to go outside the root when we compute the final product requested:

$\frac{73}{99} \,*\,7\,\sqrt{3} =\frac{511\,\,\sqrt{3} }{99}$

This is an irrational number due to the fact that it is the product of a rational number (quotient between 511 and 99) times the irrational number $\sqrt{3}$

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3 years ago