Bar graph display directly the varibales which are the rate and ratio of the numbers to visualize and display the results, these contrasting the different outcomes. On the contrary, histogram is used in grouped frequency parameters. Moreover, as little as the given five parameters or data set this will be ineffective and will result to a bar graph only and basically, the suited option is the aforementioned vertical graph to display the numbers. To expound on the definition of histogram it is used when the frequency is grouped. For example the data set of 1-5, 6-10, 11-15 and 16-20 this now can be used and applied to illustrate histogram because of the number and quantity of the given data.<span>
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Study all notes, reread the chapters again. Have someone ask questions on the chapters page by page. This always has worked for me. Plus try to do this again the night before the test. You will be surprised how much you can remember by doing it again the night before the test. Hope this helps.
f(x) = 3 - 2sin(x)
0 = 3 - 2sin(x)
- 3 - 3
-3 = -2sin(x)
-2 -2
1¹/₂ = sin(x)
sin⁻¹(1¹/₂) = sin⁻¹[sin(x)]
sin⁻¹(1¹/₂) = x
Answer:
3
Step-by-step explanation:
Answer:
y = 3.75x+5
Step-by-step explanation:
*** some of your points might be wrong ***
1. Take two points on the line. Ex: (0, 5) and (4, 20)
2. Find the slope using the change in y divided by the change in x.
Ex: m=(20 - 5)/(4 - 0) = 15/4 -> m= 3.75
3. Look at the graph for the y-intercept (when x =0)
y -intercept -> b = 5
Check:
y = mx + b; m = 3.75 (4, 20) -> x= 4 and y =20
20 = 3.75(4) + b
20 = 15 + b
5 = b
4. plug in to slope-intercept form: y = mx +b
y = 3.75x+5
