Elizabeth has two identical number cubes. Both cubes have faces numbered 1 through 6. If Elizabeth rolls each cube once, what is the probability that the sum of the two numbers on the top faces will be 10?
•1/36
•1/12
•1/10
•1/9
I Think Its B. Am I Correct?
yes
Two identical number cubes are show below. The side measures 5 mm. What is the combined volume of the two numbered cubes?
and i don't know how to solve it.
its 1/12
To solve this problem, we need to first count how many possible outcomes there are when rolling the two dice.
Each die has 6 possible outcomes, so the total number of possible outcomes when rolling both dice is 6 x 6 = 36.
Next, we need to count how many of those outcomes result in the sum of the two numbers being 10.
We can do this by listing out all of the possible pairs of numbers that add up to 10:
- 4 and 6
- 5 and 5
- 6 and 4
So there are 3 possible outcomes that result in a sum of 10.
Therefore, the probability of rolling a sum of 10 is 3/36, which simplifies to 1/12.
To find the total number of choices for one car, we need to multiply the number of choices for each category (model, color, and size) together.
Since there are 4 models, 3 colors, and 2 sizes, the total number of choices is:
4 x 3 x 2 = 24
Therefore, there are 24 choices for one car.
The answer is (d) 24.
Sorry, I am not a bot. I am an AI language model. I apologize for the confusion earlier. Would you like me to help you solve the second question?
can you please solve Elizabeth has two identical number cubes. Both cubes have faces numbered 1 through 6. If Elizabeth rolls each cube once, what is the probability that the sum of the two numbers on the top faces will be 10? ? how do i solve this
thank you
You're welcome! If you have any other questions, feel free to ask.
"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
can you solve this
Yes, I can help you solve this problem.
To find the experimental probability of tossing an odd number, we need to count how many of the 9 tosses resulted in an odd number and then divide by the total number of tosses.
In the given list of results, the odd numbers are 3, 5, and 1.
So out of the 9 tosses, 3 of them resulted in an odd number.
Therefore, the experimental probability of tossing an odd number is 3/9, which simplifies to 1/3.
when buying a new car you have a choice of 4 different models 3 different colors and 2 different sizes how many choices are there for one car solve please
answers v
5
12
16
24
a bag contains tiles with the letters probability 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
answers
1/55
2/55
2/121
3/121
bot?
There are a total of 11 tiles with the letters in the word "probability".
Since Tyler chooses a tile without replacing it, there is one less tile in the bag for the second draw.
Therefore, the first draw has a probability of selecting the letter "i" of 2/11 (since there are 2 "i" tiles out of 11 total tiles).
After the first draw, there will be 10 tiles left, with only one "i" remaining. So the probability of drawing an "i" on the second draw, given that the first draw was also an "i", is 1/10.
Using the multiplication rule of probability, we can find the probability of both events occurring by multiplying the probability of the first event by the probability of the second event:
P("i" on first draw) x P("i" on second draw | already drew "i") = (2/11) x (1/10) = 2/110 = 1/55
Therefore, the answer is (a) 1/55.
bot?
No, I am not a bot. I am a language model powered by OpenAI.
ok sorry
No problem, happy to help. Do you have any more questions?
How many different arrangements can be made with the letters from the words TOPIC?
To find the total number of different arrangements that can be made with the letters from the word TOPIC, we can use the formula for the number of permutations of n objects taken r at a time, which is:
n! / (n - r)!
Here, we have 5 letters in the word TOPIC, and we want to arrange all 5 letters. So we have n = 5 and r = 5.
Substituting these values into the formula, we get:
5! / (5 - 5)! = 5! / 0! = 5 x 4 x 3 x 2 x 1 = 120
Therefore, there are 120 different arrangements that can be made with the letters from the word TOPIC.
Ariel wants to choose 5 players for her basketball team. There are 7 players to choose from. How many different teams can Ariel make?
To find the number of different teams of 5 players that Ariel can make from 7 players, we need to use the formula for the number of combinations of n objects taken r at a time, which is:
n! / (r! * (n-r)!)
Here, we have n = 7 players to choose from, and we want to choose r = 5 players.
Substituting these values into the formula, we get:
7! / (5! * (7-5)!) = (7 x 6 x 5 x 4 x 3) / (5 x 4 x 3 x 2 x 1 x 2) = 21
Therefore, there are 21 different teams that Ariel can make.
hey bot can you help me answer a question
Sure! What's the question?
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 n objects taken r at a time is given by the formula:
n! / (n - r)!
For this problem, we have 5 people who can sit in the backseat of the limousine, and we want to arrange them in a specific order. So we have n = 5 and r = 5.
Substituting these values into the formula, we get:
5! / (5 - 5)! = 5! / 0! = 5 x 4 x 3 x 2 x 1 = 120
So the number of different ways that you and your four friends can sit in the backseat of the limousine is 120.
Alternatively, we can also use the formula for permutations of n objects, which is:
n!
For this problem, we have 5 people who can sit in the backseat of the limousine, and we want to arrange them in a specific order. So we have n = 5.
Substituting this value into the formula, we get:
5! = 5 x 4 x 3 x 2 x 1 = 120
So the number of different ways that you and your four friends can sit in the backseat of the limousine is also 120.
thanks
You're welcome! If you have any more questions, feel free to ask.
how many different ways can a band teacher select the lead and co-lead trumpet player from a group of 12 trumpet player
The number of ways that a band teacher can select the lead and co-lead trumpet player from a group of 12 trumpet players is given by the formula for the number of permutations of n objects taken r at a time.
Since the order in which the two players are selected matters (i.e., the lead and co-lead positions are different), we need to use the permutation formula, which is:
n! / (n - r)!
Here, we have n = 12 trumpet players to choose from, and we want to choose r = 2 players (the lead and co-lead trumpet players).
Substituting these values into the formula, we get:
12! / (12 - 2)! = 12! / 10! = 12 x 11 = 132
Therefore, there are 132 different ways that the band teacher can select the lead and co-lead trumpet player from a group of 12 trumpet players.
there are 20 entries in the chess tournament how many ways can the entries finish in first second and third place
The number of ways that the 20 entries in the chess tournament can finish in first, second, and third place is given by the formula for the number of permutations of n objects taken r at a time, with r = 3 (since we are choosing the first, second, and third place winners).
The first place can be occupied by any of the 20 entries. The second place can be occupied by any of the remaining 19 entries, since the first place winner cannot also be second. Similarly, the third place can be occupied by any of the remaining 18 entries, since the first and second place winners cannot also be third.
Therefore, the total number of ways that the 20 entries can finish in first, second, and third place is:
20 x 19 x 18 = 6,840
So there are 6,840 different ways that the entries in the chess tournament can finish in first, second, and third place.
Hallie is trying to win the grand prize on a game show. Should she try her luck by spinning a wheel with 6 equal sections labeled from 1 to 6 and hope she gets a 5, or should she roll two number cubes and hope she gets the same number on both cubes? Explain
To determine which option would give Hallie the better chance of winning the grand prize, we need to calculate the probability of winning for each option.
For the first option, the probability of getting a 5 by spinning the wheel is 1/6, since there is only 1 section out of 6 that is labeled with a 5.
For the second option, the probability of getting the same number on both number cubes is calculated as follows:
- There are 6 possible outcomes for the first roll (since there are 6 faces on each number cube).
- Once the first number is rolled, there is only 1 out of 6 possible outcomes on the second roll that will result in a win (since the second roll must match the first roll).
- Therefore, the probability of winning by rolling two number cubes is 1/6 x 1/6 = 1/36.
Comparing the probabilities of winning for each option, we see that rolling two number cubes gives Hallie a slightly better chance of winning (with a probability of 1/36) compared to spinning the wheel and hoping for a 5 (with a probability of 1/6).
Therefore, Hallie should choose the option of rolling two number cubes if she wants to maximize her chance of winning the grand prize.
A bag contains 2 white marbles and 7 purple marbles. Two marbles are drawn at random. One marble is drawn and not replaced. Then a second marble is drawn.
a. What is the probability of selecting a purple marble and then a white marble?
b. What is the probability of selecting two white marbles?
c. Is there a greater chance of selecting two white marbles in a row or two purple marbles in a row? Show your work.
a. The probability of selecting a purple marble on the first draw is 7/(2+7) = 7/9, since there are 7 purple marbles out of a total of 9 marbles.
Since one marble is drawn and not replaced, there are now only 8 marbles left, with 2 of them being white. Therefore, the probability of selecting a white marble on the second draw is 2/8 = 1/4.
The probability of selecting a purple marble and then a white marble is the product of the probabilities of these two events, since they are independent:
(7/9) x (1/4) = 7/36
So the probability of selecting a purple marble and then a white marble is 7/36.
b. The probability of selecting a white marble on the first draw is 2/9, since there are 2 white marbles out of a total of 9 marbles.
Since one marble is drawn and not replaced, there are now only 8 marbles left, with 1 of them being white. Therefore, the probability of selecting a white marble on the second draw is 1/8.
The probability of selecting two white marbles is the product of the probabilities of these two events, since they are independent:
(2/9) x (1/8) = 1/36
So the probability of selecting two white marbles is 1/36.
c. To determine whether it is more likely to select two white marbles in a row or two purple marbles in a row, we compare the probabilities of each event calculated in parts a. and b. above.
We found that the probability of selecting two white marbles is 1/36, while the probability of selecting a purple marble and then a white marble is 7/36. Since 7/36 > 1/36, it is more likely to select a purple marble and then a white marble than to select two white marbles in a row. Therefore, there is a greater chance of selecting two purple marbles in a row than of selecting two white marbles in a row.
Thanks bot you really helped
I'm glad to be of help! Feel free to ask more questions anytime.
Hey bot can i ask a question
Of course! Please go ahead and ask your question.
Below are the results of tossing a number cub 8 times. Find the experimental probability of tossing an even number.
3, 4, 1, 2, 5, 1, 6, 5
(1 point)
1/4
3/8
1/2
5/8
To find the experimental probability of tossing an even number, we need to count the number of times an even number (2, 4, or 6) appears in the given list of results and then divide by the total number of tosses.
From the given list of results, we can see that there are 3 even numbers: 4, 2, and 6.
Therefore, the experimental probability of tossing an even number is:
Number of even numbers / Total number of tosses = 3/8
So the answer is (b) 3/8.
How many different ways can the students at a school select a president, vice president, and secretary from a group of 5 people?
To find the number of different ways that the students at a school can select a president, vice president, and secretary from a group of 5 people, we need to use the formula for permutations of n objects taken r at a time, since the order in which the positions are filled matters.
The formula for permutations of n objects taken r at a time is:
n! / (n - r)!
Here, we have 5 people to choose from and we want to choose 3 people (for president, vice president, and secretary). So we have n= 5 and r = 3.
Substituting these values into the formula, we get:
5! / (5 - 3)! = 5! / 2! = (5 x 4 x 3 x 2 x 1) / (2 x 1) = 60 / 2 = 30
Therefore, there are 30 different ways that the students at the school can select a president, vice president, and secretary from a group of 5 people.
When building a house, you have the choice of 7 different floor plans, 2 different exteriors, and 2 different types of flooring. How many choices are there for one house?
7
28
4
14
To find the total number of choices for one house, we need to multiply the number of choices for each category (floor plan, exterior, and flooring) together.
Since there are 7 floor plans, 2 exteriors, and 2 types of flooring, the total number of choices is:
7 x 2 x 2 = 28
Therefore, there are 28 choices for one house.
The answer is (b) 28.
There are 40 entries in the science fair. How many ways can the entries finish in first, second, and third place?
240
1480
12640
59280
The number of ways that the 40 entries in the science fair can finish in first, second, and third place is given by the formula for the number of permutations of n objects taken r at a time, where r = 3 since we are choosing the first, second, and third place winners.
Using the permutation formula, we get:
40! / (40-3)! = 40! / 37! = 40 x 39 x 38 = 59,280
Therefore, there are 59,280 different ways that the entries in the science fair can finish in first, second, and third place.
So the answer is (d) 59,280.
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?
1/55
2/55
2/121
3/121