Please help me with these questions please I have 12 questions and will reward anyone who helps me with them all this is 11/12

Bingel [31]

A personal balance sheet provides an overall snapshot of your wealth at a specific period in time. It is a summary of your assets (what you own), your liabilities (what you owe) and your net worth (assets minus liabilities).

8 0

3 years ago

I need help with this please help its confusing me.

olga_2 [115]

Answer:

56 oz I think.

3 0

2 years ago

Find the value of x. <br> A. 2 <br> B. 3 <br> C. 4 <br> D. 5

masya89 [10]

The correct answer is three.

7 0

3 years ago

PLEASE HELPPPP (with steps)

yan [13]

Answer:

the answer is x=-25/56

Step-by-step explanation:

2 4/5x=-1 1/4

14/5x=-5/4

x=-25/56

7 0

3 years ago

Suppose that a fast food restaurant decides to survey its customers to gauge interest in a breakfast menu. After surveying multi

Harlamova29_29 [7]

Answer:

$ME= \frac{Width}{2}=\frac{0.078}{2}=0.039$

$\hat p =0.688+0.039=0.727$

$\hat p =0.766-0.039=0.727$

Step-by-step explanation:

Assuming that the confidence interval is (0.688; 0.766)

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".

The population proportion have the following distribution

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

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.

$z_{\alpha/2}=1.96$

Use the confidence interval to find the point estimate and margin of error for the proportion

The margin of error is given by :

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

And for our case we can find the width of the confidence interval like this:

Width =0.766-0.688=0.078

And the estimation for the margin of error would be given by:

$ME= \frac{Width}{2}=\frac{0.078}{2}=0.039$

Now we can find th point of estimate adding the margin of error to the lower limit of the interval or subtracting the margin of error to the upper limit, like this:

$\hat p =0.688+0.039=0.727$

$\hat p =0.766-0.039=0.727$

4 0

3 years ago