Where to locate the nonnegative abscissa, positive ordinate

1 answer:

4 0

The location to the nonegative abscissa, positive ordinate is in the Q1 it means it is in the fisrt quadrant as seen in the next image: http://www.mathnstuff.com/math/spoken/here/1words/q/q2.htm
Hope this helps

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Which expresion means the same as 2 1/5

Mandarinka [93]

The answer would be -10 if you want me to explain tell me to <span />

4 0

3 years ago

The water in a reservoir decreased by 47% during a drought. Before the drought, the reservoir held 44,580 acre ft (AF) of water.

RSB [31]

44,580 x 47 %

44,580 x .47 = 20,952.6

So, the water decreased 20,952.6 ft

If you want the water level after the drought just subtract.

44,580 - 20,952.6 = 23,627.4 ft

Hope this helps. :)

5 0

3 years ago

The measure of an angle is 8 degrees less than three times te measure another angle. If the two angles are supplementary, what i

tatyana61 [14]

Let's solve this problem step-by-step.

First of all, let's establish that supplementary angles are two angles which add up to 180°.
Therefore:

Equation No. 1 -
x + y = 180°

After reading the problem, we can convert it into an equation as displayed as the following:

Equation No. 2 -
3x - 8 + x = 180°

Now let's make (y) the subject in the first equation as it is only possible for (x) to be the subject in the second equation. The working out is displayed below:

Equation No. 1 -
x + y = 180°
y = 180 - x

Then, let's make (x) the subject in the second equation & solve as displayed below:

Equation No. 2 -
3x - 8 + x = 180°
4x = 180 + 8
x = 188 / 4
x = 47°

After that, substitute the value of (x) from the second equation into the first equation to obtain the value of the other angle as displayed below:

y = 180 - x
y = 180 - ( 47 )
y = 133°

We are now able to establish that the value of the two angles are as follows:

x = 47°
y = 133°

In order to determine the measure of the bigger angle, we will need to identify which of the angles is larger.

133 is greater than 47 as displayed below:

133 > 47

Therefore, the measure of the larger angle is 133°.

4 0

3 years ago

What is the measure of

siniylev [52]

Answer:

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4 0

2 years ago

Could someone help with the questions in the images below? I found them difficult

zubka84 [21]

Question 1a - You have got this correct, the median mark is 35
Question 1b- To work out the range, you must do the largest value subtract the smallest value. For your data, this would be 57 - 13 = 44

Question 2 - You have drawn the lines in the correct places, but you have not used a ruler. In order to get full marks on a question like this you must use a ruler.

Question 3 - Your lower quartile and median is correct, to find the upper quartile, we do 3(n/4). 'n' is the number of data points - in this case 120.
120/4 = 30
30 x 3 = 90
Go across the graph at 90 to see when the line is hit.
From what I can tell, the Upper Quartile is 3.

Next just find the maximum value.
Again, by the looks of it, the Maximum value is 8.5.

Now you have all the data needed for the box plot.

Min = 0
LQ = 0.8
Med = 2.1
UQ = 3
Max = 8.5

Using this information, draw a box plot like you did on the previous question.
<em>Again, I must stress that any line that you draw (unless purposely curved) must be drawn with a ruler; even on the graph at the top, during you working out. If you do not use a ruler, marks can be lost/taken away.</em>

Question 4a - For the median on a cumulative frequency chart, you must find the halfway point in the data. For this, we do n/2. In this case, n = 40
40/2 = 20
Draw a line (using ruler) across at 20 until you reach the line.
Draw another one down to help you read the number.
The median looks like 34 seconds.
Question 4b - For this question, we already know the Min, Med and Max. Now we must work out the LQ and UQ.

Remember, LQ = n/4
In this case, n = 40
40/4 = 10
Look across at ten and you will find that the LQ = 16 seconds

To work out the UQ, we multiply the LQ place by 3 3(n/4).
3 x 10 = 30
30 on the graph takes us up to 45 seconds.

We now know that the:
Min = 9
LQ = 16
Med = 34
UQ = 45
Max = 57

With this information, draw a box plot like you have in the previous questions.

Question 5 - Good things to always compare on box plots are the Min/Max/Range and the Inter Quartile Range.
The boys had a lowest minimum and a higher maximum, this means that their range is larger, resulting in a large spread of data in comparison to the girls.
The Inter Quartile Range is the difference between the Upper Quartile and the Lower Quartile (how wide the box is).
The boy IQR = 45 - 16 = 29
The girls IQR = 34 - 23 = 11
Again, the girls have much more concise results; even though they did not get the quickest result, they were more like one another.

Question 5a - The median in a box plot is always the line down the centre of the box. In this case:
Median = 24 marks
Question 5b - The IQR is always the UQ - LQ
With this data, the IQR is the following:
36 - 17 = 19 marks

Hope this helps

6 0

3 years ago