I have a practice math test and I am stuck on a few. Please help :-)
1) Write the number of permutations in factored form. Then simplify. How many different ways can you and four of your friends sit in the back of a limousine?
A) 4!; 24
B) 4!; 120
C) 5!; 120
D) 5!; 720
2) Below are the results of tossing a number cube 9 times. Find the experimental probability of tossing an odd number.
4,3,6,6,2,5,3,5,1
A) 4/9
B) 1/2
C) 5/9
D) 2/3
3) How many different ways can a band teacher select the lead and co-lead trumpet player from a group of 12 trumpet players?
A) 132
B) 264
C) 312
D) 448
4) When buying a new dress, you have a choice of 3 different lengths, 5 different colors, and 2 different styles. How How many choices are there for one dress.
A) 13
B) 15
C) 30
D) 32
5) There are 20 entries in the chess tournament. How many ways can the entries finish in in first, second, and third place?
A) 340
B) 6,840
C) 7,220
D) 8,350
6) A bag contains tiles with the letters P-R-O-B-I-L-I-T-Y. Tyler chooses a tile without looking and doesn't replace it. He chooses a second tile without looking. What is the probability that he will choose the letter I both times?
A) 1/55
B) 2/55
C) 2/121
D) 3/121
7) A sandwich shop offers the toppings shown below. How many 2-topping sandwiches can you make?
Ham
Turkey
Lettuce
Tomato
Bacon
Cheese
Mustard
A) 14
B) 21
C) 28
D) 36
its the end of the year no one cares we just wanna pass XD
This is Pro-Truth-Efficient speaking to you personally.
Here are the REAL answers
(1) Jerry mixes the letters F, P, M, M, M, F, P, and J thoroughly. Without looking, Amanda draws one letter. Expressed as a fraction, decimal, and percentage, what is the probability that J will not be the letter Amanda selects?
Answer = 7/8, 0.875, 87.5%
(2) Adam mixes the letters R, E, A, D, I, N, G, S, and A thoroughly. Without looking, Allen draws one letter. Expressed as a fraction, decimal, and percentage, what is the probability that Allen will not select a consonant?
Answer = 4/9, 0.444, 44.4%
(3) Suppose you have a drawer full of white, black, and yellow pairs of socks. If the probability of picking a white pair of socks is four-ninths, and the probability of picking a black pair of socks is
start fraction 7 over 18 end fraction, what is the probability of picking a yellow pair of socks?
Answer = 1/6
(4) The sample space for a roll of two number cubes is shown in the table.
What is the probability that the roll will result in two odd numbers?
Answer = 1/4
(5) The two numbers rolled can be added to get a sum. Find P(sum is even).
Answer = 1/2
(6) While remodeling the house, you have 3 choices of paint color, 4 choices of carpet color, and 5 choices of furniture style. How many different groupings will you be able to make using one paint color, one carpet color, and one furniture style?
Answer = 60
(7) There are four marbles in a bag with the colors red, white, blue, and green. John pulls out one marble and tosses a coin. How many elements are there in the sample space?
Answer = 8
(8) A sandwich shop sells sausage sandwiches, bacon sandwiches, and 16 different toppings. How many choices are there for a single sandwich with one topping?
Answer = 32
(9) The probability it will snow in the next two weeks is start fraction 1 over 12 end fraction for this week and one-fourth for next week. What is P(snow this week, then snow next week)?
Answer = 1/48
(10) Typically, 10% of students make a D on their tests, 60% make a C on their tests, and 30% make an A. Mrs. Smith uses a random-number table to find the experimental probability that of 5 students, at least 3 will make a C. The digit 0 represents students who make a D. The digits 1, 2, 3, 4, 5, and 6 represent students who make a C. The digits 7, 8, and 9 represent students who make an A. Find the experimental probability that of 5 students, at least 3 will make a C.
Answer = 13/20
(11) Eric has two identical number cubes. The faces of each cube are numbered 1–6. Eric will roll both of the number cubes at the same time. What is the probability that both of the numbers showing face-up will be multiples of 3?
Answer = 1/9
(12) How many different arrangements can be made with the letters from the word MATH?
Answer = 24
(13) Ariel wants to choose 5 players for her basketball team. There are 7 players to choose from. How many different teams can Ariel make?
Answer = 21
(14)Write the number of permutations in factorial form. Then simplify.
How many different ways can you and five friends sit in your assigned seats when you go to a concert?
Answer = 6!; 720
(15) Below are the results of tossing a number cube 9 times. Find the experimental probability of tossing an odd number.
4, 3, 6, 6, 2, 5, 3, 5, 1
Answer = 5/9
(16) How many different ways can a band teacher select the lead and co-lead trumpet player from a group of 12 trumpet players?
Answer = 132
(17) When buying a new dress, you have a choice of 3 different lengths, 5 different colors, and 2 different styles. How many choices are there for one dress?
ANswer = 30
(18) There are 20 entries in the chess tournament. How many ways can the entries finish in first, second, and third place?
Answer = 6,840
(19) A bag contains tiles with the letters P-R-O-B-A-B-I-L-I-T-Y. Tyler chooses a tile without looking and doesn’t replace it. He chooses a second tile without looking. What is the probability that he will choose the letter I both times?
Answer = 1/55
(20) A sandwich shop offers the following toppings. How many two-topping sandwiches can you make?
lettuce
tomato
bacon
cheese
mustard
Answer = 10
Have a nice day and DON'T forget to check out more of Pro-Truth-Efficient's answers on jiskha.com
Diana dont be annoying
diana, nobody cares, i'm trying to check my answers here you stupid cow
Al, it's too, not to.
Can someone just give answers instead of getting involved in pointless online drama that started over a math problem?
if you think cheating is wrong then why the heck are you here?Jiskha = cheating website
if you think cheating is wrong then why the heck are you here?Jiskha = chating website
You want me to give you the answers?
PS.
The test answers change in a while, so make sure TO READ THE QUESTIONS CAREFULLY!!!
Wow
Please either explain or give them :) It does not matter to me.
Well da.mn
100% thx :)
1. Is d
I’m still trying to figure out the last ones myself to
"Hon this is cheating" does that sound very nice to You,=) ?
OoOoooOOh
thx
get a clue
How many different ways can a band teacher select the lead and co-lead trumpet player from a
group of 12 trumpet players?
(1 point)
132
264
312
448
The answer is 132.
When buying a new dress, you have a choice of 3 different lengths, 5 different colors, and 2
different styles. How many choices are there for one dress?
(1 point)
13
15
30
There are 30 choices for one dress.
. There are 40 entries in the science fair. How many ways can the entries finish in first, second,
and third place?
(1 point)
240
1,480
12,640
59,280
The number of ways the science fair entries can finish in first, second, and third place is 40 x 39 x 38, which simplifies to 59,280 ways. Therefore, the answer is 59,280.
A bag contains tiles with the letters C-O-M-B-I-N-A-T-I-O-N-S. Lee chooses a tile without
looking and doesn’t replace it. He chooses a second tile without looking. What is the
probability that he will choose the letter O both times?
First, we determine the number of letters in the bag. There are 13 letters in "COMBINATIONS", so there are 13 tiles. When Lee chooses a tile the first time, there are 13 tiles to choose from. After he chooses the tile, there are only 12 tiles left. Therefore, the probability of choosing the letter O the first time is 2/13. Since he doesn't replace the first tile, there are only 12 tiles left when he chooses the second tile. One of these tiles is the letter O, so the probability of choosing that tile is 1/12.
To find the probability of both events happening, we multiply the probabilities:
P(choose O twice) = P(choose O on first draw) x P(choose O on second draw given that O was chosen on first draw)
P(choose O twice) = (2/13) x (1/12)
P(choose O twice) = 1/78
Therefore, the probability that Lee will choose the letter O both times is 1/78.
A bag contains tiles with the letters C-O-M-B-I-N-A-T-I-O-N-S. Lee chooses a tile without
looking and doesn’t replace it. He chooses a second tile without looking. What is the
probability that he will choose the letter O both times?
1/132
1/72
1/66
1/23
The probability that Lee will choose the letter O on the first draw is 2/13, since there are two O's out of a total of 13 letters in the bag.
Since he doesn't replace the first tile, there are only 12 tiles left when he chooses the second tile. One of these tiles is the letter O, so the probability of choosing that tile is 1/12.
To find the probability of both events happening, we multiply the probabilities:
P(choose O twice) = P(choose O on first draw) x P(choose O on second draw given that O was chosen on first draw)
P(choose O twice) = (2/13) x (1/12)
P(choose O twice) = 1/78
Therefore, the answer is 1/78.
there is no 1/78
I apologize for the mistake in my previous response. You are correct that 1/78 is not listed in the answer choices. The answer is actually 1/169, which is not listed in your options either.
To arrive at this answer, we first determine the number of letters in the bag. There are 13 letters in "COMBINATIONS", so there are 13 tiles. When Lee chooses a tile the first time, there are 13 tiles to choose from. After he chooses the tile, there are only 12 tiles left. Therefore, the probability of choosing the letter O the first time is 2/13. Since he doesn't replace the first tile, there are only 12 tiles left when he chooses the second tile. One of these tiles is the letter O, so the probability of choosing that tile is 1/12.
To find the probability of both events happening, we multiply the probabilities:
P(choose O twice) = P(choose O on first draw) x P(choose O on second draw given that O was chosen on first draw)
P(choose O twice) = (2/13) x (1/12)
P(choose O twice) = 1/169
Therefore, the correct answer is 1/169.
your wrong there is no 1/169 for a answer
I apologize for my previous error. The correct answer is not provided within the options given.
Write the number of permutations in factorial form. Then simplify. How many different ways can you and four of your friends sit in the backseat of a limousine?
The number of permutations of seating 5 people in a backseat can be written as 5!, which is equal to 5 x 4 x 3 x 2 x 1.
To find out how many ways you and four of your friends can sit in the backseat of a limousine, we simply substitute 5 for n in 5!, which gives us:
5! = 5 x 4 x 3 x 2 x 1 = 120
Therefore, there are 120 different ways that you and your four friends can sit in the backseat of a limousine.