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2 years ago

HELP ME PLEASEEEE HELPPPPPPPPPPPP

Mathematics

1 answer:

mario62 [17]2 years ago

7 0

Answer:

y / cos x

Step-by-step explanation:

<u>Use trigonometry to help</u>

cos = adjacent side/hypotenuse
hypotenuse = adjacent side / cos

<u>Substitute values:</u>

Distance between the Sun and the Moon = y / cos x

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X^2-18x+67=0 Solve the quadratic equation by completing the square.

Assoli18 [71]

Answer:

x = 9 ± √14

Step-by-step explanation:

x² − 18x + 67 = 0

Move the constant to the other side:

x² − 18x = -67

Take half of -18, square it, and add to both sides.

(-18/2)² = (-9)² = 81

x² − 18x + 81 = -67 + 81

x² − 18x + 81 = 14

Factor the perfect square:

(x − 9)² = 14

Solve for x:

x − 9 = ±√14

x = 9 ± √14

5 0

3 years ago

Is there anyone that can help me with this problem?

saul85 [17]

Answer:

your answer will be option 2

None of the choices are correct.

Step-by-step explanation:

have a nice day and good luck!!!

6 0

3 years ago

Read 2 more answers

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

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

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