By visual inspection, determine the best-fitting regression model for the data plot below

On a coordinate plane, a dashed straight line has a positive slope and goes through (negative 3, 1) and (0, 3). Everything to th

Svet_ta [4321]

Answer:

$y>\frac{2}{3}x+3$

Step-by-step explanation:

Step 1

Calculate the slope of the dashed line

The slope between two points is computed with the formula:

$m=\frac{y2-y1}{x2-x1}$

We consider

(-3,1) and (0,3)

and substitute into it:

$m=\frac{3-1}{0+3}$

$m=\frac{2}{3}$

Step 2

Determine the equation of the dashed line in slope-intercept form:

$y=mx+b$

The equation we established is:

$m=\frac{2}{3}$

$b=3$ ---> associated problem

Substituting yields:

$y=\frac{2}{3}x+3$

Step 3

Establish the inequality corresponding to the graph

Given that it is represented by a dashed line, the area to the left of this line is shaded.

Thus,

$y>\frac{2}{3}x+3$

Refer to the attached diagram for enhanced understanding of the problem.

8 0

9 days ago

Read 2 more answers

A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150

lawyer [4008]

Answer:

Part 1) The equation is $A(x)=150x-2x^2$

Part 2) When x=40 m, the area of the schoolyard is A=2,800 m^2

Part 3) The valid domain consists of all real numbers exceeding zero and below 75 meters

Step-by-step explanation:

Part 1) Formulate an expression for A(x)

Let

x -----> the length of the rectangular school yard

y ---> the width of the rectangular school yard

It is known that

The perimeter for the fencing (taking the school wall as one side) is

$P=2x+y$

$P=150\ m$

thus

$150=2x+y$

$y=150-2x$ -----> this is equation A

The area of the rectangular school yard is

$A=xy$ ----> this is equation B

Substituting equation A into equation B yields

$A=x(150-2x)$

$A=150x-2x^2$

Change to function notation

$A(x)=150x-2x^2$

Part 2) What is the area when x=40?

With x equal to 40 m

substitute the value into the expression from Part 1 to determine A

$A(40)=150(40)-2(40^2)$

$A(40)=2,800\ m^2$

Part 3) What would be a suitable domain for A(x) in this scenario?

We understand that

A signifies the area of the rectangular school yard

x characterizes the length of the rectangular school yard

It follows that

$A(x)=150x-2x^2$

This forms a vertical parabola opening downwards

The vertex indicates a maximum point

The x-coordinate of the vertex corresponds to the length that maximizes the area

The y-coordinate of the vertex denotes the maximum area

The vertex corresponds to (37.5, 2812.5)

Refer to the accompanying figure

Consequently,

The peak area achieved is 2,812.5 m^2

The x-intercepts are located at x=0 m and x=75 m

The domain for A is the range -----> (0, 75)

All real numbers greater than zero and less than 75 meters

5 0

11 days ago

What is the nearest ten thousand, the population of Vermont was estimated to be about 620,000 in 2008 . What might have been the

AnnZ [3877]

Answer:

The exact population of Vermont in 2008 could have been around 618,000

Step-by-step explanation:

* Here's how rounding to the nearest ten thousand works:

- Numbers ending with the last four digits between 0001 and 4999 are rounded down to the nearest lower multiple of ten thousand

- Example: 83,525 rounds down to 80,000.

- If the last four digits are 5000 or above, round up to the next higher ten thousand

- Example: 58,988 rounds up to 60,000

* Applying this rule to the problem given:

Since the rounded population is 620,000 to the nearest ten thousand, the actual population could be any value with the last four digits from 0001 to 4999 (like 618,000) or from 5000 upwards (like 624,000).

Therefore, 618,000 might represent the actual population of Vermont in 2008

6 0

15 days ago

Nico owns 11 instructional piano books. Two are beginner books, six are intermediate books, and three are advanced books. If two

PIT_PIT [3919]

There are 2 beginner, 6 intermediate, and 3 advanced books, totaling 11.

P(advanced) = 3/11
replaced
P(beginner) = 2/11

P(both) = 3/11 * 2/11 = 6/121 <=

8 0

4 days ago

Find the slope of a pipe that slopes down 2/5 inch per foot.

babunello [3635]

Answer:

$m=\frac{1}{30}$