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3 years ago

If f(x) = 3x^2+ 3x^(1/2) + 3x, then f(-9) is equal to

Mathematics

1 answer:

ivanzaharov [21]3 years ago

5 0

Since

$f(x)=3x^2+3\sqrt{x}+3x$

the domain of f(x) is the set of non-negative integers, because we can't feed a square root with negative inputs. So, you couldn't comput f(-9)

Unless you're allowed to use complex numbers, in that case we have

$f(-9)=3\cdot (-9)^2+3\sqrt{-9}+3\cdot (-9)=3\cdot 81+3\cdot (9i)-27$

Which evaluates to

$243+27i-27=216+9i$

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WILL MARK BRAINLIEST PLEASE HELP FAST !!!!! What is an equation in point-slope form for the linear function whose graph contains

liraira [26]

Answer:

y-0=7(x-5)

Step-by-step explanation:

We must first calculate the slope (m) by using the following formula:

$m=\frac{y_{2}-y_1}{x_{2}-x_{1}} =\frac{0-(-7)}{5-4} =\frac{7}{1} =7$

So our slope (m) is 7. Now we will plug in our slope and we can choose any point either (4,-7) or (5,0). I will choose (5,0) since it will make the math easier to work with and so:

$y-y_{1}=m(x-x_{1})\\\\y-(0)=7(x-5)$

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3 years ago

The interest for Kim’s loan went from 3% to 4,2 %. By how many percentage points did it increas?

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Answer: 1.2% Increase

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Step-by-step explanation:

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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

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

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