The scatter plot below shows the relationship between two variables, x and y. Which kind best fits the data?

Mathematics

1 answer:

almond37 [142]3 years ago

8 0

Answer:

The last one on the bottom right

Step-by-step explanation:

When we are trying to fit a line for a number of points, we can fix this line at several points on the graph. However, out of the several lines that we can fit, only one of these lines would work perfectly and it is called the line of best fit.

The reason why it is called the line of best fit is that it is drawn in a manner in which it leaves equal number of points above it as below it.

Supposed we have 15 points for our graph with the line of the slope passing thorough 5, the line of best fit in this case would leave 5 points above and another 5 below the line.

Leaving 6 and 4 is also manageable, but in case where 3 and 7 are left above and below the line, then it becomes a problem

Thus, out of all the representations we have, the one on the bottom right best works

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Center: (-9, -3), point on circle: (-11, 2)

Vsevolod [243]

Answer:

y = - $\frac{5}{2}$ x- $\frac{51}{2}$

Step-by-step explanation:

6 0

3 years ago

PRECAL:<br> Having trouble on this review, need some help.

ra1l [238]

1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

$\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}$

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

$\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}$

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

$\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}$

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

$\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}$

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

$\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}$

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

$\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}$

8. Divide through by x² again:

$\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}$

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

$\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}$