Answer:
140
Step-by-step explanation:
To construct a subset of S with said property, we have two choices, include 3 in the subset or include four in the subset. These events are mutually exclusive because 3 and 4 can not both be elements of the subset.
First, let's count the number of subsets that contain the element 3.
Any of such subsets has five elements, but since 3 is already an element, we only have to select four elements to complete it. The four elements must be different from 3 and 4 (3 cannot be selected twice and the condition does not allow to select 4), so there are eight elements to select from. The number of ways of doing this is
.
Now, let's count the number of subsets that contain the element 4.
4 is already an element thus we have to select other four elements . The four elements must be different from 3 and 4 (4 cannot be selected twice and the condition does not allow to select 3), so there are eight elements to select from, so this can be done in
ways.
We conclude that there are 70+70=140 required subsets of S.
Answer:
30,122
Step-by-step explanation:
This probably isn't the fastest way to solve but....
28,337x1.26%= 357
357x5=1785
28,337+1785=30,122
The 5 came from the # of years the population is increasing
I suck at explaining but I know it’s 4
A) A rational number that is between 9.5 and 9.7 could be 9.6 because it can be written as a fraction (in this case 9 6/10 or 9 3/5)
B) An irrational number could be the square root of 91 because it's value cannot be written as a fraction, as with all other square roots of non-perfect squares. sqrt(91) is about 9.54 rounded to the nearest hundredth.
Hope this helped!
In a parallelogram, opposite sides are parallel and equal . So to prove JKLM a parallelogram, we need to satisfy both criterias . And only the last option stand up on both criterias that is opposite sides are parallel and equal .
So the correct option is the last option .