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3 years ago

Which of the following best defines 2 to the power of 2 over 3?

Mathematics

2 answers:

EleoNora [17]3 years ago

4 0

Hello,

$2^{\frac{2}{3}}=\sqrt[3]{2^2} =\sqrt[3]{4}$

Answer A

Arisa [49]3 years ago

4 0

The correct answer is the first option. Two to the power of 2 over 3 is best defined by Cube root of 4. It is best understood when written in numbers.

2^(2/3)

The denominator here represents the radical and the numerator is the power by which the base is raised. Therefore, it can be written as:

∛2^2 or ∛4

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How can you solve this problem 2250=5000 x R x 5

vesna_86 [32]

R is equal to 450

first you subtract 5000 from each side

-2250=Rx5
then divid by 5

R=450

5 0

3 years ago

F(3)= 5x^2 - 4x + 9 evaluate this quadratic function

Neporo4naja [7]

Answer:

$f(3) = 38$

Step-by-step explanation:

Given the quadratic function

$f\left(x\right)=\:5x^2\:-\:4x\:+\:9$

Evaluating the quadratic function

$f\left(x\right)=\:5x^2\:-\:4x\:+\:9$

substitute x = 3

$f\left(3\right)=\:5\left(3\right)^2\:-\:4\left(4\right)\:+\:9$

$f(3) = 45-16+9$

$f(3) = 38$

Therefore,

$f(3) = 38$

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2 years ago

The picture explains the question

liubo4ka [24]

Answer:

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4 0

3 years ago

Read 2 more answers

–24 + 12d = 2(d – 3) + 22

iris [78.8K]

Answer:

d = 4

Step-by-step explanation:

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3 years ago

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Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

 $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

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3 years ago