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2 years ago

Which expression is equivalent to 3 cube root (343x^9 y^12 z^6) ? 7x3y4z2 7x3y6z2 49x3y6z2 49x3y4z2

2 answers:

7 0

Cube root of 343 = 7

of x^9 it is x^3 ( because x^9 = x^3 * x^3 * x^3)

of y^12 it is y^4

of z^6 it is z^2

So the correct choice is A

3241004551 [841]2 years ago

3 0

Answer:

A. $7x^3*y^4*z^2$

Step-by-step explanation:

We have been given an expression $\sqrt[3]{343x^9y^{12}z^6}$ . We are asked to find the equivalent expression to our given expression.

Using exponent property $(a^b)^c=a^{b\cdot c}$ we can rewrite the terms of our given expression as:

$\sqrt[3]{7^3*(x^3)^3*(y^4)^{3}*(z^2)^3}$

Using property $\sqrt[n]{a^n}=a$ we will get,

$7x^3*y^4*z^2$

Therefore, the simplified form of our given expression would be $7x^3*y^4*z^2$ and option A is the correct choice.

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What number is 0.72 more than 0.971?

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Answer:

1.691

Step-by-step explanation:

.971 + .72

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2 years ago

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Consider the following sample of observations on coating thickness for low-viscosity paint.

Julli [10]

Answer:

a) $\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And for this case if we use this formula we got:

$\bar x = 1.3538$

b) Since we have n =16 values for the sample the median can be calculated as the average between position 8th anf 9th and we got:

$Median = \frac{1.31+1.46}{2}= 1.385$

c) $P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=1.3538 +1.28*0.3505=1.8024$

So the value of height that separates the bottom 90% of data from the top 10% is 1.8024.

d) $Median= \frac{x_{8} +x_{9}}{2}$

The variance for this estimator is given by:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} Var(X_{8} +X_{9})$

We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

And the standard error would be given by:

$Sd(\frac{x_{8} +x_{9}}{2})= \sqrt{0.0482}=0.2196$

Step-by-step explanation:

Data given:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31 1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83

Part a

We can calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And for this case if we use this formula we got:

$\bar x = 1.3538$

Part b

For this case in order to calculate the median we need to put the data on increasing way like this:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31 1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83

Since we have n =16 values for the sample the median can be calculated as the average between position 8th anf 9th and we got:

$Median = \frac{1.31+1.46}{2}= 1.385$

Part c

For this case we can assume that the mean is $\mu = 1.3538$

And we can calculate the population deviation with the following formula:

$\sigma = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{N}}$

And if we replace we got: $\sigma= 0.3105$

And assuming normal distribution we have this:

$X \sim N (\mu = 1.3538, \sigma= 0.3105)$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=1.3538 +1.28*0.3505=1.8024$

So the value of height that separates the bottom 90% of data from the top 10% is 1.8024.

Part d

The median is defined as :

$Median= \frac{x_{8} +x_{9}}{2}$

The variance for this estimator is given by:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} Var(X_{8} +X_{9})$

We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

And the standard error would be given by:

$Sd(\frac{x_{8} +x_{9}}{2})= \sqrt{0.0482}=0.2196$

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Tricia got a 6% raise on her weekly salary. The raise was $30 per week. What was her original salary?

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motikmotik

I assume there is a missing piece of information here, either the area or the perimeter should be given.
I'll just tell you how to solve this type of problems and you can apply using the information in your question.

If YV is the width of the rectangle, then YX will be its length.

Now, if the area is known:
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Answer:

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Step-by-step explanation:

if there is 8 shelves that display 1 car then that means there is 8 cars total.

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(12-8) which would equal 4.

you need 4 cars left but on each shelf going forward needs to have 2 cars on it.

now since you have 4 cars you would divide that by the amount of cars on each shelf going forward.

4/2) which equals 2.

you need 2 more shelfs for the 4 cars needed.

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