Please help with these problems!!!

Hey is anyone from flvs school 7 grade Florida please i need help do u know the questions for the DBA just give me the questions

tatiyna

Hey I'm in flvs too but I have a different teacher, what class and module is your DBA for?

4 0

3 years ago

at the beginning of the month a store had a balance of -554. during the month the store lost another 600 what is the balance

blondinia [14]

Answer:

-554 - 600 = -1154

the store's balance is $ -1154.00

3 0

3 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.02}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2$

$n = 2401$

The minimum sample size is 2401.

4 0

3 years ago

The following set of numbers represents the number of hours a group of students spent reading over the course of two weeks.

irina1246 [14]

Answer:

i don't know. sorry!

Step-by-step explanation:

5 0

2 years ago

Please answer as many as you can.

dimaraw [331]

Answer: okay so your problems are like 10000000k through out resons of dum. so x=-1. is a little wrong becayse the x = into a 1 so you make the 1 that factor. so thats why answer for 2.5

Step-by-step explanation:

6 0

3 years ago