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3 years ago

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Tell if 2, 3, 6 & 9 are factors of the given number, give a partner for each trick factor. make a list of factors.<br /&g

t;<br />A. 3,141<br />B. 57<br />C. 52,24,382<br />D. 1,440<br />E. 82<br />F. 5,126,848
I NEED ANSWERS ASAP

I NEED ANSWERS ASAP NOW!

1 answer:

oksian1 [2.3K]3 years ago

3 0

If you give me answers i can help you

You might be interested in

41>6m-7 help please

oksian1 [2.3K]

41> 6m-7.
So, I added 7 to both sides and got 48>6m and then divided both sides by 6 to get 8>m or m<8.

3 0

3 years ago

Most US adults have social ties with a large number of people, including friends, family, co-workers, and other acquaintances. I

xxTIMURxx [149]

Answer:

$t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971$

$p_v =2*P(t_{(1699)}>1.971)=0.049$

If we compare the p value and the significance level given $\alpha=0.1$ 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 true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=669$ represent the sample mean

$s=732$ represent the sample standard deviation

$n=1700$ sample size

$\mu_o =68$ represent the value that we want to test

$\alpha=0.$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:

Null hypothesis: $\mu = 634$

Alternative hypothesis: $\mu \neq 634$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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".

Calculate the statistic

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

$t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=1700-1=1699$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(1699)}>1.971)=0.049$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ 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 true mean is different from 634 at 10% of signficance.

3 0

3 years ago

Using the given points and line, determine the slope<br><br> (0,32) and (100,212)

suter [353]

To solve for the slope given two lines, use the formula:

(y₂ - y₁)
----------
(x₂ - x₁)

Set one of the points as (x₁, y₁), and the other as (x₂, y₂).

(x₁, y₁) = <span>(0,32)
</span>(x₂, y₂) <span>= (100,212)

plug into corresponding places:

</span>(y₂ - y₁) (212 - 32) (180)
---------- = -------------- = -------
(x₂ - x₁) (100 - 0) (100)

180/100 is your slope

If you want simplified, it will be: 9/5

hope this helps

5 0

3 years ago

Ben weighs 140.8 pounds, and Jennifer weighs 117.7 pounds. How many more pounds does Ben weigh than Jennifer?

DENIUS [597]

Answer:

Ben weighs 23.1 lbs more than Jennifer.

Step-by-step explanation:

To find the difference between Ben and Jennifer's weight, you have to subtract the smaller number from the bigger number. In this case, subtract Jennifer's weight from Ben's weight. This would look like this: 140.8-117.8.

140.8-117.7=23.1

Therefore, Ben weighs 23.1 lbs more than Jennifer.

8 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago