In order to put two quantities as a relationship, just put a colon or a fraction bar in between them, in the order that the problem asks you.
In this problem, it's asking for the # of students planning to go to an instate college to the # of students planning to go to an out of state college.
So, 120 : 70
Even though it doesn't tell you to simplify, it will never hurt you, so always do it.
Both 120 and 70 can be divided by 10, so the simplified answer would be 12:7
Hope this helps!
<span>These are five questions and five answers.
Part 1.] Identify the following series as arithmetic, geometric, both, or neither. Includes picture
Answer: arithmetic
Justification:
The arithmetic series are those in which the distance (difference) between any consecutive terms is constant, so you can find the term An adding the difference to the previous term, A(n-1).
In this case the series is:
∞
∑ (3 + na) =>
n=1
Term1: 3 + a
Term2: 3 + 2a
Term3: 3 + 3a
Term3: 3 + 4a
As you see, the difference between two consecutive terms is a, which shows that it is an arithmetic series.
Part 2.] Evaluate C(3, 3)
A.] 1
B.] 6
C.] 24<span>
Answer: option A) 1.
Explanation:
The formula for combinations is Cm,n = m! / [n! (m - n)! ]
So, for C3,3, C3,3 = 3! / [3! 0!] = 3! / 3! = 1.
From this you might learn that Cm,m for any m is always 1.
Part 3.] Evaluate C(6,3)
A.] 20
B.] 120
C.] 720
Answer: option A) 20
Explanation:
Use the same formula, Cm,n = m! / [n! (m - n)! ]
C6,3 = 6! / [3! 3! ] = 6*5*4 / (3*2*1) = 5*4 = 20
Part 4.] </span>How many subsets of four elements each exist in a set of seven elements?
A.] 4!
B.] C(7, 4)
C.] P(7, 4)
Answer: option B) C(7,4)
Explanation
You have to select collections of 4 elements from a set of 7 elements, and must realize that the order is irrelevant, i.e. it is the same the set ABCD as the set ACBD or DCAB or any combination of those four elements. So, the answer is a combination and not a permutation.
The number of subsets of 4 elements that you can form with 7 elements is: C(7,4), which is the option B.
Part 5.] What is the probability of having 3 children that are all boys?
A.] 1/2
B.] 1/8
C.] 3/8
Answer: option B: 1/8
Explanation:
The number of possible outcomes is 2 * 2 * 2 = 8
Only one of those outcomes is for all boys = 1
Therefore the probability is 1/8
You can see that representing the possible outcomes using B for boys and G for girls:
1) BBB (three boys)
2) BBG
3) BGB
4) BGG
5) GBB
6) GBG
7) GGB
8) GGG
Those are the 8 possible outcomes and as seen only one is for 3 boys.
</span>
19 guests because each guest is $2 each according to the question.
It should be 8 if it is 5-5 which gives you 0.
Or it could be (5x5)-5=20+8=28.
