How do you find the range of a linear graph

the ratio of the number of adults to the number of students at the prom has to be 1:10 lat year there were 477 more students tha

maksim [4K]

You would divide 10 into 477 to get 47.7, but you can't put 47.7 because you can't have a decimal. The answer is 47

8 0

2 years ago

A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the

EastWind [94]

Answer:

a) $\hat p=\frac{471}{1024}=0.460$

The standard error is given by:

$SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156$

And the margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305$

b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

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

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

Data given and notation

n=1024 represent the random sample taken

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

$\hat p=\frac{471}{1024}=0.460$ estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

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.

$z_{\alpha/2}=1.96$

The standard error is given by:

$SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156$

And the margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305$

Part b

If we replace the values obtained we got:

$0.460-1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.429$

$0.460+1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.491$

The 99% confidence interval would be given by (0.429;0.491)

8 0

2 years ago

How do you solve 15x1.12

Andru [333]

Put 1.12 on top and times regullarly you should get 1280.00

6 0

3 years ago

Read 2 more answers

Solve the system of linear equations. separate the x- and y- values with a coma. -8x=72-16y

serious [3.7K]

Let's use the "elimination by addition/subtraction" method to solve this system for x and y.

 -8x=72-16y
4x=-92+16y <= Mult. all 3 terms by 2: 8x = - 184 + 32y

Now combine the following, column by column:

 -8x= 72-16y
8x = - 184 + 32y
-----------------------
0 = -112 + 16y. Solving for y, 16y = 112, and y = 7.

Now subst. 7 for y in either of the original equations and calculate x:

4x=-92+16(7) => 4x = -92 + 112 = 20. Then x = 5.

Solution is 5, -7 (or, better written) (5, - 7).

8 0

3 years ago

Enter the equation of the line in slope-intercept form. Enter the answer in fraction form.

Montano1993 [528]

Answer:

11/2??????

Step-by-step explanation:

6 0

2 years ago