Help. Your backyard is 10 feet longer than it is wide. Its area is 8624 square feet. Its width is Solve in quadratic equation.

What is 4 3/10 plus (-6.2) in decimal form

k0ka [10]

Answer:

Step-by-step explanation:

First change 4 3/10 to 4.3. Then you change 4.3 + -6.2 to 4.3 - 6.2.

Then you do 4.3 - 6.2 and you should get - 1.9.

5 0

3 years ago

Read 2 more answers

Can someone please explain how to solve such a mean and sneaky problem? Please?

Lelu [443]

Answer:

all real numbers

Step-by-step explanation:

simplify each side of the inequality

26 + 6b -- already fully simplified

2(3b + 4) → 6b + 8

26 + 6b ≥ 6b + 8

subtract 6b from both sides

26 ≥ 8

since this expression is true (26 is greater than 8), then the solution is all real numbers. any value could take the place of b because in the end, it will be canceled out by itself, resulting in 26 ≥ 8

4 0

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.