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rusak2 [61]
3 years ago
10

Which expression is equivalent to ^3 square root x^10

Mathematics
2 answers:
Luden [163]3 years ago
5 0

Answer:

D

Step-by-step explanation:

The second part is C x^3 3 square root x

The third part is a = -8 b= -2

jeka943 years ago
3 0

Answer:

D. 3√x^9 • x

Step-by-step explanation:

3√x^10

A. 3√3x³ + x

B. 3√3x³ * x

= 3√3x^4

C. 3√x³ + x³ + x³ + x

= 3√3x³ + x

D. 3√x^9 • x

= 3√x^(9 + 1)

= 3√x^10

The equivalent expression to 3√x^10 is 3√x^9 • x

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D 100cm

Step-by-step explanation:

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In a machine, a large gear completes a revolution every minute while a small gear completes a revolution every 24 seconds. If th
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The hiking trail 2600 miles long and passes through fourteen states. Because it is their first time hiking the trail, Janet and
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Answer:

They will hike 16% of the trail in Georgia.

Step-by-step explanation:

We have that:

The hiking trail is of 2600 miles.

416 of those miles are in Georgia?

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Georgia distance multiplied by 100% and divided by the total distance. So

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6 0
2 years ago
Simplify to create an equivalent expression.<br> (4+2k)⋅6+3k =
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The answer is
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8 0
3 years ago
According to a recent​ study, some experts believe that 15​% of all freshwater fish in a particular country have such high level
Lera25 [3.4K]

Answer:

The approximate probability that the market will have a proportion of fish with dangerously high levels of mercury that is more than two standard errors above 0.15 is 0.95.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 \mu_{\hat p}=0.15

The standard deviation of this sampling distribution of sample proportion is:

 \sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}

As the sample size is large, i.e. <em>n</em> = 150 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by the normal distribution.

Compute the mean and standard deviation as follows:

\mu_{\hat p}=0.15\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.15(1-0.15)}{150}}=0.0292

So, \hat p\sim N(0.15, 0.0292^{2})

In statistics, the 68–95–99.7 rule, also recognized as the empirical rule, is a shortcut used to recall that 68%, 95% and 99.7% of the Normal distribution lie within one, two and three standard deviations of the mean, respectively.

Then,

                                  P (µ-σ < X < µ+σ) ≈ 0.68

                                  P (µ-2σ <X < µ+2σ) ≈ 0.95

                                  P (µ-3σ <X < µ+3σ) ≈ 0.997

Then the approximate probability that the market will have a proportion of fish with dangerously high levels of mercury that is more than two standard errors above 0.15 is 0.95.

That is:

P(\mu_{\hat p}-2\sigma_{\hat p}

4 0
3 years ago
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