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Pavel [41]
3 years ago
15

HOW DO YOU SOLVE THIS PROBLEM

Mathematics
1 answer:
Elenna [48]3 years ago
3 0

x = 100° (using definition of vertical angles)

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Help with all please
seropon [69]

a) First, draw a graph. The x axis should be numbered: 0, 1, 2, 3, 4, 5, 6, 7, 8. Each number is the last digit of the year. For each year, there is one average score that has to be plotted. Number the y axis of the graph from 0-600, by increments of 50, or something around there. Now, with each year (each x value) place the dot as high as the average score. Repeat with all years. DO NOT draw a line to connect them. A scatter plot is a bunch of points that are not connected.

b) Use the form (y2-y1)/(x2-x1) where (x1,y1) (x2,y2) are any two points from 2001-2006. It does not matter which points, pick any, and assign each of the numbers x1,y1 and x2,y2. Plug it in to the equation, and simplify. This is your slope, also called m. Now plug m into the equation y-y1=m(x-x1) where y1 and x1 are the x and y coordinates of any point from 2001 to 2006. x and y dont have values themselves, they stay there. Now distribute the m into the parenthesis, add y1 to both sides (to cancel it out on the left), and you should be left with the same equation, but now in y=mx+b form. B is where the line crosses the y axis, so put a point on the vertical axis where b is. M is your slope, so every time you go to the right one number, go up M numbers. Draw another point. Repeat this until you can connect these new dots into a line. This line should be on the same graph as your other points, but might not touch all of the scatter plot points. That's still okay, just leave it as is.

c) Find the point where x=6 (meaning the year is 2006) on your line. Find the average test score for that year by seeing where the point lines up along the y axis (draw a straight line from the point to the left to see, if you have to). Take that number and add 1m (the m is the number you used in step B) to it. This predicts what the average score might be after 1 (that's why the 1 is there) year. This is a predicted value, and might not be perfectly correct!! Write/record how close this new, predicted, score is to the actual number, 515, which can be found on the scatter plot.

d) repeat step C exactly as before (still use the year 2006) , but add 4m instead of 1m, to get the predicted score 4 years after 2006 (2010), instead of 1 year after (2007).

I hope I have been of some help! Best of luck!!! :)

3 0
3 years ago
A recipe needs 1/4 tablespoon salt.
-Dominant- [34]

Answer:

stan LOONA

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The vertices of a swimming pool are E(5, 5), F(5, 20), G(35, 20), and H(35, 5). The endpoints of a line that divides the pool in
sleet_krkn [62]

Answer:

Step-by-step explanation:

Download pdf
3 0
3 years ago
Y= -4x cos(x) -5
saw5 [17]
To be honest I do t understand this but Carlos does so contact him
3 0
3 years ago
Pls help I don't understand:
sineoko [7]

Answer:

Part a) T(d)=2d+30

Part b) T(6)=42\ minutes

Step-by-step explanation:

Part a) Write an equation for T (d)

Let

d ----> the number of days

T ---> the time in minutes of the treadmill

we know that

The linear equation in slope intercept form is equal to

T=md+b

where

m is the slope or unit rate

b is the y-intercept or initial value

In this problem we have

The slope or unit  rate is

m=2\ \frac{minutes}{day}

The y-intercept or initial value is

b=30\ minutes

substitute

T(d)=2d+30

Part b) Find T (6), the minutes he will spend on the treadmill on day 6

For d=6

substitute in the equation and solve for T

T(6)=2(6)+30

T(6)=42\ minutes

8 0
3 years ago
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