jerry mixes the letters f p m m m f p and j thoroughly. without looking Amanda draws one letter. expressed as a decimal and percentage, what is the probability that j will not be the letter Amanda selects. i need answers bad at math
8 letters, and 7 of them are not j
so, what do you think?
i still need help tho
Sure, do you have any other questions or do you need further explanation about the previous question?
2. Christopher mixes the letters S, E, L, E, C, T, E, and D 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? (1 point)%0D%0A%09, 0.375, 37.5%%0D%0A%09, 0.6, 60%%0D%0A%09, 0.625, 62.5%%0D%0A%09, 0.5, 50%
We know that a consonant is any letter that is not a vowel. In this case, the vowels in the set are E. Thus, the consonants are S, L, C, T, and D.
The total number of letters in the set is 8. The number of consonants is 5. So, the number of vowels in the set is 3.
The probability of Allen selecting a consonant is the number of consonants divided by the total number of letters:
P(consonant) = 5/8
The probability of Allen not selecting a consonant is the number of vowels divided by the total number of letters:
P(not consonant) = 3/8
Expressed as a fraction, the probability that Allen will not select a consonant is 3/8.
To express this as a decimal, we divide the numerator by the denominator:
3 ÷ 8 = 0.375
So, the decimal form of the probability is 0.375.
To express this as a percentage, we multiply the decimal by 100:
0.375 × 100 = 37.5%
Therefore, the probability that Allen will not select a consonant is 3/8, 0.375, or 37.5%.
and can you give the answer without the explanation to not take up so much space on here please?
The answer is:
P(not consonant) = 3/8, 0.375, or 37.5%
3.%C2%A0%C2%A0The sample space for a roll of two number cubes is shown in the table. %C2%A0(1,1) %C2%A0(1,2) %C2%A0(1,3) %C2%A0(1,4) %C2%A0(1,5) %C2%A0(1,6) %C2%A0(2,1) %C2%A0(2,2) %C2%A0(2,3) %C2%A0(2,4) %C2%A0(2,5) %C2%A0(2,6) %C2%A0(3,1) %C2%A0(3,2) %C2%A0(3,3) %C2%A0(3,4) %C2%A0(3,5) %C2%A0(3,6) %C2%A0(4,1) %C2%A0(4,2) %C2%A0(4,3) %C2%A0(4,4) %C2%A0(4,5) %C2%A0(4,6) %C2%A0(5,1) %C2%A0(5,2) %C2%A0(5,3) %C2%A0(5,4) %C2%A0(5,5) %C2%A0(5,6) %C2%A0(6,1) %C2%A0(6,2) %C2%A0(6,3) %C2%A0(6,4) %C2%A0(6,5) %C2%A0(6,6) What is the probability that the roll will result in both numbers being the same? %C2%A0(1%C2%A0point)
There are 36 total outcomes (6 outcomes for the first cube multiplied by 6 outcomes for the second cube).
There are 6 outcomes where both cubes are the same (1,1; 2,2; 3,3; 4,4; 5,5; 6,6).
Therefore, the probability of rolling both numbers the same is:
6/36 = 1/6
So, the probability that the roll will result in both numbers being the same is 1/6, which is approximately 0.1667 or 16.67%.
4. The sample space for a roll of two number cubes is shown in the table.%0D%0A%0D%0A (1,1)%09 (1,2)%09 (1,3)%09 (1,4)%09 (1,5)%09 (1,6)%0D%0A (2,1)%09 (2,2)%09 (2,3)%09 (2,4)%09 (2,5)%09 (2,6)%0D%0A (3,1)%09 (3,2)%09 (3,3)%09 (3,4)%09 (3,5)%09 (3,6)%0D%0A (4,1)%09 (4,2)%09 (4,3)%09 (4,4)%09 (4,5)%09 (4,6)%0D%0A (5,1)%09 (5,2)%09 (5,3)%09 (5,4)%09 (5,5)%09 (5,6)%0D%0A (6,1)%09 (6,2)%09 (6,3)%09 (6,4)%09 (6,5)%09 (6,6)%0D%0A%0D%0AThe two numbers rolled can be added to get a sum. Find P(sum is less than 4). (1 point)%0D%0A%09%0D%0A%09%0D%0A%09%0D%0A%09%0D%0A5. You are packing for a vacation. You pack 3 pairs of shorts, 5 shirts, and 2 pairs of shoes. How many different outfits will you be able to make using one pair of shorts, one shirt, and one pair of shoes? (1 point)%0D%0A%0910%0D%0A%0917%0D%0A%0930%0D%0A%0935%0D%0A6. A spinner has 3 equal sections: red, white, and blue. John spins the spinner and tosses a coin. Which shows the sample space for spinning the spinner and tossing the coin? (1 point)%0D%0A%09%0D%0A %09 Red%09 White%09 Blue%0D%0A Heads%09 Heads, Red%09 Heads, White%09 Heads, Blue%0D%0A Tails%09 Heads, Red%09 Heads, White%09 Heads, Blue%0D%0A%09%0D%0A %09 Red%09 White%0D%0A Heads%09 Heads, Red%09 Heads, White%0D%0A Tails%09 Heads, Red%09 Heads, White%0D%0A%09%0D%0A %09 Red%09 White%09 Blue%0D%0A Heads%09 Heads, Red%09 Heads, White%09 Heads, Blue%0D%0A%09%0D%0A %09 Red%09 White%09 Blue%0D%0A Heads%09 Heads, Red%09 Heads, White%09 Heads, Blue%0D%0A Tails%09 Tails, Red%09 Tails, White%09 Tails, Blue%0D%0A7. A yogurt shop offers 2 different flavors of yogurt and 11 different toppings. How many choices are there for a single serving of yogurt with one topping? (1 point)%0D%0A%0920%0D%0A%0911%0D%0A%0913%0D%0A%0922%0D%0A8. The probability it will snow in the next two weeks is for this week and for next week. What is P(snow this week, then snow next week)? (1 point)%0D%0A%09%0D%0A%09%0D%0A%09%0D%0A%09%0D%0A9. Suppose 10% of the flights arriving at an airport arrive early, 60% arrive on time, and 30% arrive late. Valerie used the random-number table to find the experimental probability that of 5 flights, at least 2 will arrive late. The digit 0 represents flights arriving early. The digits 1, 2, 3, 4, 5, and 6 represent flights arriving on time. The digits 7, 8, and 9 represent flights arriving late.%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0AFind the experimental probability that of 5 flights, at least 2 will arrive late. (1 point)%0D%0A%09%0D%0A%09%0D%0A%09%0D%0A%09%0D%0A10. How many different arrangements can be made with the letters from the word MATH? (1 point)%0D%0A%0912%0D%0A%0924%0D%0A%09106%0D%0A%091,210%0D%0A11. Ariel wants to choose 5 players for her basketball team. There are 7 players to choose from. How many different teams can Ariel make? (1 point)%0D%0A%0921%0D%0A%0935%0D%0A%0942%0D%0A%0956%0D%0A12. Write the number of permutations in factorial form. Then simplify. How many different ways can you and six of your friends sit in your assigned seats in math class? (1 point)%0D%0A%096!; 120%0D%0A%096!; 720%0D%0A%097!; 2,520%0D%0A%097!; 5,040%0D%0A13. Below are the results of tossing a number cube 7 times. Find the experimental probability of tossing an even number.%0D%0A%0D%0A6 4 3 2 5 3 3 (1 point)%0D%0A%09%0D%0A%09%0D%0A%09%0D%0A%09%0D%0A14. How many different ways can a coach select the captain and co-captain of a team from a group of 20 people? (1 point)%0D%0A%0940%0D%0A%09160%0D%0A%09380%0D%0A%09420%0D%0A15. 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)%0D%0A%0913%0D%0A%0915%0D%0A%0930%0D%0A%0932%0D%0A16. There are 20 entries in the chess tournament. How many ways can they finish in first, second, and third place? (1 point)%0D%0A%09340%0D%0A%096,840%0D%0A%097,220%0D%0A%098,350%0D%0A17. 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 point)%0D%0A%09%0D%0A%09%0D%0A%09%0D%0A%09Start Fraction 1 over 23 End Fraction%0D%0A18. An ice cream shop offers the toppings shown below. How many 4-topping ice cream sundaes can you make?%0D%0A%0D%0Achocolate chips%0D%0Awalnuts%0D%0Astrawberry sprinkles%0D%0Acaramel%0D%0Ahot fudge%0D%0Awhipped cream%0D%0Agummy bears (1 point)%0D%0A%0911%0D%0A%0928%0D%0A%0930%0D%0A%0935
4. The sum can be less than 4 if we roll (1,1), (1,2), or (2,1), so there are 3 possible rolls that result in a sum less than 4.
Thus, the probability of rolling a sum less than 4 is:
3/36 = 1/12
Therefore, the probability that the sum is less than 4 is 1/12, which is approximately 0.0833 or 8.33%.
5. To find the number of outfits we can make, we need to multiply the number of choices for each item together:
3 pairs of shorts x 5 shirts x 2 pairs of shoes = 30 outfits
Therefore, we can make 30 different outfits using one pair of shorts, one shirt, and one pair of shoes.
7. We have 2 choices for the flavor of yogurt, and we can add any one of 11 toppings. Thus, there are:
2 x 11 = 22
choices for a single serving of yogurt with one topping.
9. To find the experimental probability that at least 2 flights out of 5 will arrive late, we need to count the number of outcomes where this occurs. The digits 7, 8, and 9 represent flights arriving late, so we need to count the number of times that at least 2 of the digits are 7, 8, or 9 in a sequence of 5 digits.
Using the random-number table, we obtain the following sequence of digits: 0 9 2 3 7. There are 2 flights arriving late in this sequence. We continue to generate 4 more sequences and count how many have at least 2 digits that are 7, 8, or 9. Let n represent the number of sequences that have at least 2 digits that are 7, 8, or 9. Then:
n/5 = experimental probability of at least 2 flights out of 5 arriving late
Using the random-number table, we obtain the following sequences:
0 4 2 3 6 (0 flights arriving late)
5 4 6 1 9 (2 flights arriving late)
1 2 1 6 0 (0 flights arriving late)
2 5 5 0 4 (1 flight arriving late)
8 5 3 1 9 (2 flights arriving late)
The number of sequences with at least 2 digits that are 7, 8, or 9 is 2 (the second and last sequences). Therefore, the experimental probability of at least 2 flights out of 5 arriving late is:
n/5 = 2/5
10. To find the number of different arrangements that can be made with the letters in the word MATH, we need to use permutations. There are 4 letters, so there are:
4! = 4 x 3 x 2 x 1 = 24
different arrangements.
12. The number of permutations of 7 people in assigned seats is:
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040
13. The experimental probability of tossing an even number is the number of even numbers that were tossed divided by the total number of tosses:
Experimental Probability = Number of Even Tosses / Total Number of Tosses
In this case, there are 4 even numbers (2, 4, 6, and 6) in 7 total tosses, so:
Experimental Probability = 4/7
14. To find the number of ways a coach can select a captain and co-captain from a group of 20 people, we need to use permutations. There are 20 people, and we want to select 2 people in a specific order (captain first, then co-captain). Thus, there are:
20 P 2 = 20 x 19 = 380
ways to select a captain and co-captain.
15. To find the number of choices for one dress, we need to multiply the number of choices for each category together:
3 lengths x 5 colors x 2 styles = 30 choices
Therefore, there are 30 choices for one dress.
16. We need to count the number of ways the top 3 places can be awarded to the 20 entries. This is a permutation problem. There are 20 choices for first place, 19 remaining choices for second place, and 18 remaining choices for third place (once the first and second place winners are chosen). Therefore, there are:
20 P 3 = 20 x 19 x