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3 years ago

8

How many variable terms are in the expression 3x to the third power Y plus 5X to the second power plus Y +9

1 answer:

andrew-mc [135]3 years ago

8 0

Well, variables are stuff like x and y because they can represent a number. They are variable in an equation. so if you have (3x^3)+(5X^2)+Y+9 then your variables are x X and Y (im assuming the capital X is there and not a typo) if that was a typo and meant to be lowercase, then its just a 2 variable with the x and y

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In 2015 as part of the General Social Survey, 1289 randomly selected American adults responded to this question:

Radda [10]

Answer:

B (0.312, 0.364)

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 interval $1-\alpha$ , 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 $1 - \frac{\alpha}{2}$

For this problem, we have that:

1289 randomly selected American adults responded to this question. This means that $n = 1289$ .

Of the respondents, 436 replied that America is doing about the right amount. This means that $\pi = \frac{436}{1289} = 0.3382$ .

Determine a 95% confidence interval for the proportion of all the registered voters who will vote for the Republican Party. 

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3382 - 1.96\sqrt{\frac{0.3382*0.6618}{1289}} = 0.312$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3382 + 1.96\sqrt{\frac{0.3382*0.6618}{1289}} = 0.364$

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3 years ago

Write the formula for the parabola that has x- intercepts (3,0) and (8,0), and y- intercept (0,−2)

alexira [117]

Answer: $x^2-5x=-12(y+2)$

Step-by-step explanation:

Given

Parabola has x-intercept has $(3,0)$ and $(8,0)$

and Y-intercept as $(0,-2)$

Now the general equation of parabola is

$y=ax^2+bx+c\quad \ldots(i)$

Substitute $(0,-2)$ in $(i)$ we get

$-2=c$

Now substitute $(3,0)$ in equation $(i)$

$0=a(3)^2+3b-2\quad \ldots(ii)$

Now substitute $(8,0)$ in equation $(i)$

$0=a(8)^2+8b-2\quad \ldots(iii)$

Solving $(ii)$ and $(iii)$ we get

$a=\frac{-1}{12}\ \text{and}\ b=\frac{5}{12}$

therefore

$y=\frac{-x^2}{12}+\frac{5x}{12}-2$

$12y=-x^2+5x-24$

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4 years ago