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saul85 [17]
4 years ago
5

How do I graph this?

Mathematics
1 answer:
Mice21 [21]4 years ago
4 0
In a scatterplot you add a point for every data entry and do not connect them via lines or anything else.

You also labeled you axis already correct, so you pretty much only have to decide on the scale you want to use for the y-axis/number of offices.
You can choose between a relative scale similar to the years, starting at some point and ending at some or an absolute one which starts at 0.
The advantage of the first one is the "zoomed in" view, showing differences more clearly the other doesn't warp the absolute size.

For an absolute scale decide on the end (because it starts with 0) and then divide that range by the amount of squares to get your step height. I used 80000 as the end which resulted in steps of 4000 with each square.

With the relative scale its similar, round the max&min values of the data and decide on some step width (by dividing by your vertical graph height and rounding again). I used 3000 with 80000 and 20000 as the end and start of the axis.

Once you have your scale and both axis are labeled you only have to draw the points.

I added two pictures what the result should look like.

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The perimeter of a triangle is 60 feet. one side is 12 feet long. of the two unknown sides, one of them is twice as long as the
Lady bird [3.3K]
One is 36 the other is 12
6 0
3 years ago
What are the coordinates of the image of the point (2, –6) under a dilation with a center of (0, 0) and a scale factor of 1/2
ycow [4]

That would be (1, -3) because you basically divide the coordinates by the scale factor, in this case 2.

8 0
3 years ago
Hey does any body know how i can approximate 13 like this had this homework but it got late on me.​
Illusion [34]

Answer:

(see attachment)

To approximate the square root of 13:

Working from the top down...

Enter the number you are trying to approximate in the top box: \boxed{\sf \sqrt{13}}

Find the perfect squares directly below and above 13.

Perfect squares:  1, 4, 9, 16, 25, 36, ...

Therefore, the perfect squares below and above 13 are: 9 and 16

Enter these with square root signs in the next two boxes: \boxed{\sf \sqrt{9}} and  \boxed{\sf \sqrt{16}}  

Carry out the operation and enter  \boxed{\sf 3} and \boxed{\sf 4} in the next two boxes.

Enter the number you are trying to square root (13) in the top left box, the perfect square above it (16) in the box below, then the perfect square below it (9) in the two boxes to the right of these. Carry out the subtractions and place the numbers in the boxes to the right.

\dfrac{\boxed{\sf 13}-\boxed{\sf 9}}{\boxed{\sf 16}-\boxed{\sf 9}}=\dfrac{\boxed{\sf 4}}{\boxed{\sf 7}}

Now enter the number you are trying to square root (13) under the square root sign.  Place the square root of the perfect square below it (3) in the box to the right.  Copy the fraction from above (4/7).  Finally, enter this mixed number into a calculator and round to the nearest hundredth.

\sf \sqrt{13}=\boxed{\sf3}\dfrac{\boxed{\sf 4}}{\boxed{\sf 7}}=\boxed{\sf3.57}

7 0
2 years ago
How many radians does the minute hand of a clock rotate through over 10 minutes? How many degrees?
Leto [7]

Answer:

10 minutes =  1.047198 radians

10 minutes = 60 degree

Step-by-step explanation:

we know that

There is 60 minutes in an hour

so it is This is 2 Pi radians

so 10 minutes is  \frac{10}{60} or  \frac{1}{6} th of an hour

and here we know  the hand will travel  of the full circle

and we know A full circle measures 2 Pi radians

2 × pi radians = 360 degrees

10 minutes = \frac{1}{6} ×   2 × pi radians

10 minutes =  \frac{1}{3} × pi radians

10 minutes =  1.047198 radians

10 minutes = 60 degree

3 0
3 years ago
Drag each tile to the correct box.
igor_vitrenko [27]
The answer to this statement is 2
3 0
3 years ago
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