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3 years ago

What is the answer for this problem in the picture

Mathematics

2 answers:

sdas [7]3 years ago

8 0

Answer:

The correct answer is third option

8x⁴yz⁴√2yz

Step-by-step explanation:

It is given that,

√128x⁸y³z⁹

To find the equivalent of √128x⁸y³z⁹

We have 128 = 2⁷

x⁸ = (x⁴)², y³ = y² * y and z⁹ = (z⁴)² * z

√128x⁸y³z⁹ = √(2⁷ * (x⁴)² * y²y * (z)⁴)²z)

= 2³x⁴yz⁴√(2 * y * z)

= 8x⁴yz⁴√2yz

Therefore the correct answer is third option

DENIUS [597]3 years ago

4 0

Answer:

The correct answer option is $= 8 x ^ 4 y z ^ 4 \sqrt{2yz}$ .

Step-by-step explanation:

We are given the following expression and we are to simplify it and determine whether which of the given expressions is equivalent to it, assuming $y\geq 0$ and $z\geq 0$ :

$\sqrt { 128 x ^ 8 y ^ 3 z ^ 9 }$

Breaking the terms to get:

$\sqrt { 128 x ^ 8 y ^ 3 z ^ 9 }$ $=8\sqrt{2yz} \times x^{8\times\frac{1}{2} } \times y^{2\times \frac{1}{2} } \times z^{8\times \frac{1}{2} }$

$= 8 x ^ 4 y z ^ 4 \sqrt{2yz}$

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Read 2 more answers

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oksano4ka [1.4K]

Answer:

Null hypothesis: $\mu \leq 8000$

Alternative hypothesis: $\mu > 8000$

$z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$

$p_v =P(Z>2)=0.0228$

Step-by-step explanation:

1) Data given and notation

$\bar X=8300$ represent the sample mean

$\sigma=1200$ represent the population standard deviation

$n=64$ sample size

$\mu_o =800$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 8000, the system of hypothesis are :

Null hypothesis: $\mu \leq 8000$

Alternative hypothesis: $\mu > 8000$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$

4) P-value

Since is a one-side upper test the p value would given by:

$p_v =P(Z>2)=0.0228$

5) Conclusion

If we compare the p value and the significance level assumed, for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than $8000.

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