What are some factors of 12
2 answers:
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Answer:
The most appropriate graphical display for this case is the histogram.
Step-by-step explanation:
One of the most used graphical display is <em>the histogram</em>, a term used for the first time by Karl Pearson in 1895.
Suppose the researcher measures different times for a sample of children to "<em>complete a basic reading skills task</em>". The researcher can register any of these different times for each child in the sample and then create different intervals of the same size (in seconds) for all the cases in the sample. Any of these intervals represent a <em>class</em>. For example, we can have a class for 0 to 20 seconds, other for 20 to 40 seconds, and so on.
Well, into any appropriate class, <em>we summarize the number of cases</em> that fall into it, which represents the frequency of them for this class. We can then represent in the x-axis the different intervals of time (classes) and the y-axis the frequency for each class (rectangles of different heights according to the frequency).
In doing so, we can "extract" valuable information from this graph such as (but not limited to):
- Type of distribution that probably follows the data (for instance, normal).
- The interval (or class) with the highest/lowest frequency (maybe the most probable/less probable for this sample).
- The presence of an outlier (roughly, a very extreme value).
- We can divide each frequency by the total of the values to obtain the <em>relative frequency</em>. In doing so, we can determine, for example, below which value are 60% of the cases, 10% of the cases, and so on.
- The symmetry of the graph (perfectly symmetrical or skewed).
Thus, for these and other advantages, <em>the histogram</em> is "the most appropriate graphical display for summarizing the frequency of children falling into different intervals of time".
We can show an example of a histogram with hypothetical data regarding this case (see the graph) below). For instance, the most frequent interval is the first (10 - 20) seconds with nine (9) cases, there is probably an outlier between (110 - 120) seconds and the data is not symmetrical (skewed right).
Part 1
Let:
y be the total cost
x be the number of tickets
a be fair admission
Since the fair charges $1.25 per ticket, the equation of the line would look like:
y = 1.25x + a
Since Johnny spent $43.75 and bought 25 tickets, we find a by replacing y with 43.75 and x with 25:
43.75 = 1.25×25 + a
Now solve for a
43.75 = 31.25 + a
43.75 - 31.25 = a
a = 12.50
Going back to the first equation:
y = 1.25x + 12.50
Part 2:
a) Let (9,8) be (x1,y1) and (4,-12) be (x2,y2)
The formula for slope is:
b) For the point-slope equation, I'll use (9,8)
Using the formula:
c) For the slope-intercept form, we solve for y in the point-slope formula:
Let:
y be the total cost
x be the number of tickets
a be fair admission
Since the fair charges $1.25 per ticket, the equation of the line would look like:
y = 1.25x + a
Since Johnny spent $43.75 and bought 25 tickets, we find a by replacing y with 43.75 and x with 25:
43.75 = 1.25×25 + a
Now solve for a
43.75 = 31.25 + a
43.75 - 31.25 = a
a = 12.50
Going back to the first equation:
y = 1.25x + 12.50
Part 2:
a) Let (9,8) be (x1,y1) and (4,-12) be (x2,y2)
The formula for slope is:
b) For the point-slope equation, I'll use (9,8)
Using the formula:
c) For the slope-intercept form, we solve for y in the point-slope formula: