Steps: 1. put the numbers in numerical order under the sequence it's easier
2. median is the easiest the number that's in the middle (same amount of numbers on each side. if there is an even amount of number than two will technically be in the middle of that happens add those two numbers together then divide by 2 that's your median (ex: 2, 4, 8, 7, 10, -2, 15 put them in order from least to greatest or greatest to least -2, 2, 4, 7, 8, 10, 15 . 7 would be your median. however if the sequence had one more number -2, 2, 3, 4, 7, 8, 10, 15 4 and 7 are in the middle. so 7+4= 11. 11÷2= 5.5 that's your median)
3. to find mean: add all the numbers up and decide by the amount of numbers there are. (ex: 12, 15, 2, -7. there are 4 numbers so 12+15+2+-7= 22. 22÷4= 5.5 that would be your mode)
4. to find the mode: the number that most occurs (ex: 2, 7, 13 ,2 ,-7, 4, 2. the number 2 is in that sequence 3 times that's more than any number that's your answer.)
5. to find range: find the difference between the largest and the smallest number in the sequence. (ex: 3, 4, 6, 8, 9, 16. 16 is the largest while 3 is the smallest 16-3= 13 that's you're range.
Explanation: We know that: 1 mg is equivalent to 0.0001 dag To convert 38600 mg to dag, we will simply use cross multiplication as follows: 1 mg ..............> 0.0001 dag 38600 mg ......> n
From the graph, it is obvious that the trend is decreasing from 100 on day 2, to 1 on day 10. So, the answer could either be A or C. The question would be how fast is it decreasing? To illustrate this, let's find the difference of consecutive data:
100 - 26 = 74 26 - 6 = 20 6 - 2=4 2-1=1
It must not be an additive rate because there is no common difference. Let's illustrate if the trend is in multiplicative rate:
100/26 = 3.85 26/6 = 4.33 6/2 = 3 2/1 = 2
More or less, they have a common divider. Hence, the decreasing rate is in multiplicative rate. The answer is A.
An outlier in a set of data is the one that is significantly lower or higher than the average of the totals. Whichever number skews the average is the outlier.
