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3 years ago

What is the solution to this system?

Mathematics

1 answer:

katovenus [111]3 years ago

6 0

The solution is where the two lines meet. From the the graph, it looks like
(1, -1)

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If the bakers percentage for raisins in a raisin bread formula is 8% and the bakers percentage for flour is 100%, what amount of

a_sh-v [17]

How many pounds were the 8% formula?

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3 years ago

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A 3-pound tub of butter at n dollars per pound costs $3.85. 3n = 3.85 3 - n = 3.85 3 + n = 3.85 3/n = 3.85

Elis [28]

Given:
3-pound tub of butter costs $3.85

3.85/3 = 1.2833 per pound.

3n = 3.85

3(1.2833) = 3.85
3.8499 = 3.85
3.85 = 3.85

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3 years ago

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If m // n and n // o and p is a transversal, find the

MakcuM [25]

Answer:

See below.

Step-by-step explanation:

Here is one pair of angle of each type. There are more correct answers.

corresponding angles: <K, <O

alternate interior angles: <N, <P

alternate exterior angles: <I, <S

same-side interior angles: <N, <Q

vertical angles: <P, <R

supplementary angles: <J, <M

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2 years ago

What is the vertex of f(x) = |5x| 2?

Sveta_85 [38]

The Vertex of f(x)=|5x|+2 is (0,2)
and vertex of f(x)=|5x|-2 is (0,-2)

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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

juin [17]

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.

$z_{\alpha/2}=2.58$

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%)

5 0

3 years ago