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3 years ago

Help with all of the questions?

Mathematics

1 answer:

jasenka [17]3 years ago

4 0

1. x + 3y = 3 and x - 2y = 0
2. (2,1)
3. solution
4. true
5. equation 1 : equation 2 :
x + y = 3 x - 2y = 0
2 + 1 = 3 2 - 2(1) = 0
3 = 3 (correct) 2 - 2 = 0
0 = 0 (correct)

both equations are true...so (2,1) is a solution set to this system of equations

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5x+4y = -14<br> 3x+6y=6<br> What is the answer

iren [92.7K]

Hallo! Her er svaret ditt på dette spørsmålet! ☺

Answer: If you are solving by Substitution, your answer will be...

Point form : (-6,4)

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3 years ago

Carl buys 1.5 pounds of grapes. The

Umnica [9.8K]

Answer:

$3.51

Step-by-step explanation:

1 pound is $2.39 so divide by 2 and round up to the nearest whole number which is $1.12. Add $1.12(The cost of 0.5 pounds) to $2.39(The cost of 1 pound) to get $3.51 as your total.

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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

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

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