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IceJOKER [234]
3 years ago
6

Which exponential functions have been simplified correctly check all that apply

Mathematics
2 answers:
Lesechka [4]3 years ago
9 0

Answer:

Option A, C, D, are the correct options.

Step-by-step explanation:

To check the exponential functions which have been simplified correctly we will take option one by one.

A). f(x)=5(\sqrt[3]{16})^{x}=5(\sqrt[3]{2^{4}})^{x}=5(2\sqrt[3]{2})^{x}

B). f(x)=2.3(8^{\frac{1}{2}})^{x}=2.3(2.\sqrt{2})^{x}\neq 2.3(4^{x})

C). f(x)=81^{\frac{x}{4}}=(3^{4})^{\frac{x}{4}}=(3^{\frac{4}{4}})^{x}=3^{x}

D). f(x)=\frac{3}{4}(\sqrt{27})^{x}=\frac{3}{4}(\sqrt{3^{3}})^{x}=\frac{3}{4}.(3\sqrt{3})^{x}

f(x)=(24)^{\frac{1}{3}x}=(2^{3}.3)^{\frac{1}{3}x}=(2.3^{\frac{1}{3}})^{x}\neq 2(\sqrt[3]{3})^{x}

Fittoniya [83]3 years ago
5 0
Principle: Law of Exponents - Combination of product to a power & power to a power. The first is when raising a product of two integers to a power, the power is distributed to each factor. In equation it is,

(xy)^a = (x^a)(y^a)

The latter is when raising the base with a power to a power, the base will remain the same and the powers will be multiplied. In equation it is, 
(x^a)(x^b) = x^ab

Check:


f(x) = 5*(16)^.33x = 5*(8*2)^0.33x = 5*(8^0.33x)(2^0.33x) = 5*(2^x)*(2^0.33x) = 5*(2^1.33x)

f(x) = 2.3*(8^0.5x) = 2.3*(4*2)^0.5x = 2.3*(2^x)(2^0.5x) = 2.3*(2^1.5x)

f(x) = 81^0.25x = 3^x

f(x) = 0.75*(9*3)^0.5x = 0.75*(3^x)*(3^0.5x) = 0.75*3^1.5x

f(x) = 24^0.33x = (8*3)^0.33x = (2^x)*(3^0.33x)


Therefore, the answer is third equation.

<em>ANSWER: f(x) = 81^0.25x = 3^x</em>
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To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

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The home team therefore wins 50% of its games

This means that p = 0.5

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Sample of 80 means that n = 80 and, by the Central Limit Theorem:

\mu = p = 0.65

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5*0.5}{80}} = 0.0559

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Z = \frac{X - \mu}{\sigma}

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