1)An ice cream store has 31 flavors of ice cream and 10 toppings. A regular sundae has 1 flavor of ice cream, 1 topping, and comes with or without whipped cream. How many different ice cream sundaes can be ordered?
A)310
B)372
C)620
D)82
I chose C
2)How many 5 digit codes are possible if 0 cannot be used and no digit can be repeated?
A)15,120
B)45
C)30,240
D)59,049
I chose B
3)A clown has 7 balloons, each a different color. There are 5 children. how many ways can the clown give each child a balloon?
A)21
B)5040
C)42
D)2520
I chose B
4)Evaluate C(13,9)
A)17,160
B)715
C)259,459,200
D)117
I chose D
5)The probabilty that an event will occur is 11/15. What are the odds that the event will occur?
A)15:11
B)11:15
C)4:1
D)11:4
I chose B
These questions are not about algebra. They are about probability.
1) Correct: 31x10x2
2) There are 9 choices for the first digit, 8 for the second, etc. The answer is 9*8*7*6*5 = 9!/4!
You chose incorrectly. Were you just guessing?
3) Wrong. 7*6*5*4*3 = ?
4) No. What did you choose D? A wild guess perhaps? I suggest you review the definition of "Combinations" and use the appropriate formula.
5) Wrong again. When stating odds, the ratio of the first number to the sum of both numbers is the probability the first event will occur.
Here is a popular web site about combinations and permutations. I suggest you review it, and use it for question 4)
http://mathforum.org/dr.math/faq/faq.comb.perm.html
thanks for the help. none were wild guesses
I don't understand what you said for #5 but its either C or D.
4 B) 5 D)
1) To find the number of different ice cream sundaes that can be ordered, you need to multiply the number of choices for each component together.
There are 31 flavors of ice cream to choose from, 10 toppings, and two options for whipped cream (with or without). So the total number of combinations would be 31 * 10 * 2 = 620.
Therefore, the correct answer is C) 620.
2) In this case, we need to find the number of 5-digit codes where 0 cannot be used and no digit can be repeated.
Since 0 cannot be used, we have 9 options for the first digit (1-9). For the second digit, since no digit can be repeated, we have 8 options left. Similarly, for the third digit, we have 7 options, and for the fourth and fifth digits, we have 6 and 5 options respectively.
To find the total number of codes, we multiply these options together: 9 * 8 * 7 * 6 * 5 = 15,120.
Therefore, the correct answer is A) 15,120.
3) In this scenario, we need to find the number of ways the clown can give each child a balloon, assuming each child gets one balloon.
For the first child, the clown has 7 options to choose from (any of the 7 balloons). Similarly, for the second child, there are 6 options left, then 5 for the third child, 4 for the fourth child, and 3 for the last child.
To find the total number of ways, we multiply these options together: 7 * 6 * 5 * 4 * 3 = 5040.
Therefore, the correct answer is B) 5040.
4) The expression C(13,9) represents the number of combinations of choosing 9 items from a set of 13 items.
The formula for calculating combinations is: C(n, r) = n! / (r!(n-r)!)
Plugging in the given values, we have: C(13,9) = 13! / (9!(13-9)!) = 13! / (9!4!)
Simplifying further, 13! can be written as 13 * 12 * 11 * 10 * 9!, and 4! = 4 * 3 * 2 * 1.
So the expression becomes: (13 * 12 * 11 * 10 * 9!) / (9! * 4 * 3 * 2 * 1)
The 9! terms cancel each other out, leaving us with: (13 * 12 * 11 * 10) / (4 * 3 * 2 * 1) = 17,160.
Therefore, the correct answer is A) 17,160.
5) To calculate the odds of an event, we use the formula:
Odds in favor = probability of event occurring / probability of event not occurring
Given that the probability of the event occurring is 11/15, the probability of it not occurring is 1 - 11/15 = 4/15.
Using the odds formula, we have: Odds in favor = (11/15) / (4/15) = 11/4.
Therefore, the correct answer is D) 11:4.