Module 6 What is the line of best fit? Why do we want the sum of the residuals to be as close to zero as possible?

Margaret [11]

Answer:

25%

Step-by-step explanation:

Percentages are one of several ways of describing quantities' relationships to one another. Specifying one number as a percentage of another means specifying the fraction of the second quantity the first comprises. The percentage value is the number that, divided by 100, equals that fraction. To express the percentage as a whole number, round it accordingly. Some applications, however, don't require percentages as exact whole figures.

Divide the first number the second. For instance, if you want to find what percentage 43 is out of 57, divide 43 by 57 to get 0.754386.

6 0

3 years ago

A cost regularly priced at $188 is on sale for 25% off . What is the sale price ?

Brums [2.3K]

You need to multiply 188 by 25%. But first convert 25% to a decimal 0.25.
188*0.25 is 47
Now we subtract 188 - 47
Which is 141
$141 is the sales price.

5 0

3 years ago

Read 2 more answers

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals: