john's first 3 tests scores in chemistry were 70, 75, 91. what was his median score? a. 72.5 b. 91 c. 75 d. 10
1 answer:
You might be interested in
The first thing to do is to sort the given data from smallest to largest. Doing so leads to this ordered list:
{13, 15, 15, 17, 18, 18, 20, 21, 21, 23, 26, 27, 28, 30, 32, 33, 37, 40}
From this list, we can see which values are between 10 and 19 (inclusive). Those values are {13, 15, 15, 17, 18, 18} which represents 6 values. So the first bar (for the "10-19" group) is going to be 6 units high.
The next group, in the "20-29" group, is this set of values {20, 21, 21, 23, 26, 27, 28} which is 7 items this time. The second bar is 7 units high.
The "30-39" bar will be 4 units high because there are 4 items in the set {30, 32, 33, 37}
There is only one person who is between 40 and 49, and that person is 40 years old. So the "40-49" group has a bar that has height of 1.
The final histogram and frequency chart is shown in the attached image. The frequency chart is optional, but it helps supplement the histogram in my opinion.
{13, 15, 15, 17, 18, 18, 20, 21, 21, 23, 26, 27, 28, 30, 32, 33, 37, 40}
From this list, we can see which values are between 10 and 19 (inclusive). Those values are {13, 15, 15, 17, 18, 18} which represents 6 values. So the first bar (for the "10-19" group) is going to be 6 units high.
The next group, in the "20-29" group, is this set of values {20, 21, 21, 23, 26, 27, 28} which is 7 items this time. The second bar is 7 units high.
The "30-39" bar will be 4 units high because there are 4 items in the set {30, 32, 33, 37}
There is only one person who is between 40 and 49, and that person is 40 years old. So the "40-49" group has a bar that has height of 1.
The final histogram and frequency chart is shown in the attached image. The frequency chart is optional, but it helps supplement the histogram in my opinion.
You have several possible points . Any point that lies on the given line will have a
value that you obtain by plugging in the corresponding value of
into the equation of the line.
The line has equation , so any point on the line will have the form
.
Among the four given points, there are only 2 different values of to check. We have
which means the given line contains the points (4, -5) and (1, 4). So the answer is A.