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3 years ago

Simplify. -10m+2m^4-13m-20m^4

1 answer:

8 0

First one:

you can add -10m and -13m but you can't add -10m and 2m^4 becuase the powers aren't the same so

when adding the like terms
look at the:
powers, (x^3 adds with x^3)
placehloder letter (x adds with x and y adds with y and so on)

-10m+2m^4-13m-20m^4
powers: m^1 and M^4
placeholders: all m

add
-10m-13m+2m^4-20m^4
-23m-18m^4

second one:
when multiplying exponents, you add with like
so if you multipliy
x^2yz^3 times x^4y^2z^2 thne you would get x^6y^3z^5
when multiply with coeficients
2x^2yz^3 times 4x^4y^2z^2=8x^6y^3z^5
so using associative property a(bc)=(ab)c
2/3 times p^4 times y^3 times y^4 times s^5 times 6 times p^2 times s^3
group like terms

(2/3 times 6) times (p^4 times p^2) times (y^3 times y^4) times (s^5 times s^3)
(4) times (p^6) times (y^7) times (s^8)
4p^6y^7s^8

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2 years ago

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3 years ago

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3 years ago

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

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Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.