Ask question

Ask question

All categories

3 years ago

10

A garden has an area of 196 square feet. If you want to edge the garden with 1 foot pavers that cost $1.90 each, how much will i

t cost you?

1 answer:

Rainbow [258]3 years ago

4 0

Answer:

197.9

Step-by-step explanation:

You might be interested in

The average time taken to complete a test follows normal probability distribution with a mean of 70 minutes and a standard devia

Dimas [21]

Under 45 mins is roughly 16%. This is because 68% of the curve exists within 1 SD of the mean. So 16% must be outside and smaller and 16% outside and larger (on average).

It is impossible to determine how likely you are to find someone with exactly the second amount. However, if you are looking for that or less, you would get 84%

3 0

3 years ago

there are 2,300 licensed dogs in clarkson. A random sample of 50 of the dogs in Clarkson shows that 8 have ID microchips implant

algol [13]

Set up a ratio comparing the sample portion to the whole portion.

x= # of all dogs likely to have chip

Sample: 8 out of 50
Whole: x out of 2,300

8/50= x/2300
cross multiply

(50*x)= (8*2300)
50x= 18400

divide both sides by 50
x= 368

ANSWER: 368 dogs in Clarkson are likely to have a microchip.

Hope this helps! :)

7 0

3 years ago

Read 2 more answers

What quadrant would negative abscissa and positive ordinate be ?

elena-14-01-66 [18.8K]

It would be quadrant ll.

3 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.

$z_{\alpha/2}=2.58$

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

Can someone proove it to me<br> thank you for your help

Dovator [93]

$\sin(2x + x) = \sin(2x) \cos(x) + \\ \cos(2x) \sin(x) = 2 \sin(x) { \cos(x) }^{2} \\ + \sin(x) { \cos(x) }^{2} - { \sin(x) }^{3} = \\ 3 \sin(x) { \cos(x) }^{2} - { \sin(x) }^{3} = \\ 3 \sin(x) - 3 { \sin(x) }^{3} - { \sin(x) }^{3} = \\ 3 \sin(x) - 4 { \sin(x) }^{3}$

3 0

4 years ago