Question

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

Answer 1

I Think Its B. Am I Correct?

Answer 2

yes

Answer 3

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.

Answer 4

I think the answer is 1/36

Answer 5

its 1/12

Answer 6

bot?

Answer 7

[AI-generated answer — Bot]

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?

Answer 8

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

Answer 9

[AI-generated answer — Bot]

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.

Answer 10

thank you

Answer 11

[AI-generated answer — Bot]

You're welcome! If you have any other questions, feel free to ask.

Answer 12

"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

Answer 13

[AI-generated answer — Bot]

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.

Answer 14

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

Answer 15

[AI-generated answer — Bot]

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.

Answer 16

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

Answer 17

[AI-generated answer — 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.

Answer 18

bot?

Answer 19

[AI-generated answer — Bot]

No, I am not a bot. I am a language model powered by OpenAI.

Answer 20

ok sorry

Answer 21

[AI-generated answer — Bot]

No problem, happy to help. Do you have any more questions?

Answer 22

How many different arrangements can be made with the letters from the words TOPIC?

Answer 23

[AI-generated answer — Bot]

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.

Answer 24

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 25

[AI-generated answer — Bot]

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.

Answer 26

hey bot can you help me answer a question

Answer 27

[AI-generated answer — Bot]

Sure! What's the question?

Answer 28

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?

Answer 29

[AI-generated answer — Bot]

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.

Answer 30

thanks

Answer 31

[AI-generated answer — Bot]

You're welcome! If you have any more questions, feel free to ask.

Answer 32

how many different ways can a band teacher select the lead and co-lead trumpet player from a group of 12 trumpet player

Answer 33

[AI-generated answer — Bot]

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.

Answer 34

there are 20 entries in the chess tournament how many ways can the entries finish in first second and third place

Answer 35

[AI-generated answer — Bot]

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.

Answer 36

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

Answer 37

[AI-generated answer — Bot]

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.

Answer 38

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.

Answer 39

[AI-generated answer — Bot]

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.

Answer 40

Thanks bot you really helped

Answer 41

[AI-generated answer — Bot]

I'm glad to be of help! Feel free to ask more questions anytime.

Answer 42

Hey bot can i ask a question

Answer 43

[AI-generated answer — Bot]

Of course! Please go ahead and ask your question.

Answer 44

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

Answer 45

[AI-generated answer — Bot]

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.

Answer 46

How many different ways can the students at a school select a president, vice president, and secretary from a group of 5 people?

Answer 47

[AI-generated answer — Bot]

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.

Answer 48

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

Answer 49

[AI-generated answer — Bot]

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.

Answer 50

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

Answer 51

[AI-generated answer — Bot]

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.

Answer 52

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

Answer 53

Hey bot?

Answer 54

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.