I'll just show you how to make a frequency table using the above data.
We will group the data into class intervals and determine the frequency of the group.
<span>8 12 25 32 45 50 62 73 80 99 4 18 9 39 36 67 33
</span>
smallest data value = 4
highest data value = 99
difference = 99 - 4 = 95
number of data = 17
Let us assign a class interval of 20.
Class Interval Tally Frequency
0-20 8, 12, 4, 18, 9, 5
21-40 25, 32, 39, 36, 33 5
41-60 45, 50, 67 3
61-80 62, 73, 80 3
81-100 99 1
That is how a frequency table look like. Usually, under the Tally column, tick marks are written instead of the numbers but for easier monitoring, I used the numbers in the data set.
Answer:
21/26 percent chance and if you count Y as a vowel 20/26
A number 1/10 as great as 7962 would be 796.2
You get this answer by either dividing 7,962 with 10 which gives you 769.2
Or you can multiply 7,962 with 1/10 to get 7,962/10. If you simplify then you get 796 1/5 or 796.2.
To solve the problem, get the
percentage of each test by multiplying the score and the percentage then add it all up:
82 * .25 (highest test grade) + 65* .15 (lowest test grade) +
71*.20 (each test remaining) + 77*.20 (each test remaining) + 92*.20 (homework
grade)
= 20.5 + 9.75 + 14.2 + 15.4 + 18.4 = 78.25 or 78% in whole number
Actually there is enough information to solve this
problem. First, let us find the total per row and per column.
(see attached pic)
P(Grade 10 | opposed) with P(opposed | Grade 10)
P(Grade 10 | opposed) = Number in Grade 10 who are opposed
/ Total number of Opposed (column)
P(Grade 10 | opposed) = 13 / 41 = 0.3171
P(opposed | Grade 10) = Number in Grade 10 who are opposed
/ Total number in Grade 10 (row)
P(opposed | Grade 10) = 13 / 32 = 0.4063
Therefore:
P(Grade 10 | opposed) IS NOT EQUAL P(opposed | Grade 10),
hence they are dependent events.
Answer:
P(Grade 10 | opposed) < P(opposed | Grade 10)