Drink Survey
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Drink | Number of Shoppers Who Preferred It
A | 10
B | 15
C | 7
D | 3
E | 6
What is the probability that 1 shopper, selected at random, preferred neither Drink E nor Drink C?
13/41
13/28
35/41
28/41
I also need help with this
A standard number cube is rolled 288 times. Predict how many times a 3 or a 5 will be the result.
96 times
93 times
99 times
58 times
Drink Survey
---------------------
Drink | Number of Shoppers Who Preferred It
A | 10
B | 15
C | 7
D | 3
E | 6
What is the probability that 1 shopper, selected at random, preferred neither Drink E nor Drink C?
13/41
13/28
35/41
28/41
What is the probability that neither tile is purple?
The probability of drawing a non-purple tile on the first draw is:
= (number of non-purple tiles) / (total number of tiles)
= (4 + 6) / (4 + 6 + 10)
= 10/20
= 1/2
Since the first tile is not returned, the number of tiles remaining in the box for the second draw will depend on the color of the first tile drawn. If the first tile drawn is non-purple, there will be 9 purple tiles and a total of 19 tiles for the second draw. If the first tile drawn is purple, there will be 8 purple tiles and a total of 18 tiles for the second draw.
The probability of drawing a non-purple tile on the second draw, given that the first tile drawn was non-purple, is:
= (number of remaining non-purple tiles) / (total number of remaining tiles)
= (4 + 6 - 1) / (19 - 1)
= 9/18
= 1/2
The probability of drawing a non-purple tile on the second draw, given that the first tile drawn was purple, is:
= (number of remaining non-purple tiles) / (total number of remaining tiles)
= (4 + 6) / (18 - 1)
= 10/17
The combined probability of drawing two non-purple tiles is:
= (probability of non-purple on first draw) * (probability of non-purple on second draw, given first draw was non-purple) + (probability of purple on first draw) * (probability of non-purple on second draw, given first draw was purple)
= (1/2) * (1/2) + (10/20) * (10/17)
= 1/4 + 25/68
= 47/136
Therefore, the probability that neither tile is purple is 47/136.
13/41
A 3 or 5 will occur 1/3 of the time.
i wish i could help but im bad at math
wow, you are smart PsyDAG
Thanks
Drink Survey
---------------------
Drink | Number of Shoppers Who Preferred It
A | 10
B | 15
C | 7
D | 3
E | 6
What is the probability that 1 shopper, selected at random, preferred neither Drink E nor Drink C?
13/41
13/28
35/41
28/41
To find the probability that a shopper preferred neither Drink E nor Drink C, we need to first find the total number of shoppers who preferred drinks other than E and C.
Total number of shoppers who preferred drinks other than E and C = Number of shoppers who preferred A + Number of shoppers who preferred B + Number of shoppers who preferred D
= 10 + 15 + 3
= 28
Therefore, the probability that 1 shopper, selected at random, preferred neither Drink E nor Drink C is:
Number of shoppers who preferred neither E nor C/Total number of shoppers = 28/41
So, the answer is 28/41.
he righ
Yes, that's correct.
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I'm an AI language model created to help with tasks such as answering questions and providing information. Is there anything else you need help with?
A box contains 4 yellow tiles, 6 green tiles, and 10 purple tiles. Without looking, you draw out a tile and then draw out a second tile without returning the first tile.
A farmer examines a sample of 25 cartons of eggs and finds that 3 contain cracked eggs. What is the best prediction of the number of cartons with cracked eggs in a delivery with 500 cartons?
To predict the number of cartons with cracked eggs in a delivery of 500 cartons, we can use proportion.
The proportion of cartons with cracked eggs in the sample is:
3/25
We can use this proportion to predict the number of cartons with cracked eggs in a delivery with 500 cartons by setting up a proportion:
3/25 = x/500
where x is the number of cartons with cracked eggs in a delivery of 500 cartons.
To solve for x, we can cross-multiply:
25x = 3 * 500
Simplifying:
25x = 1500
Dividing both sides by 25:
x = 60
Therefore, the best prediction of the number of cartons with cracked eggs in a delivery with 500 cartons is 60 cartons.
A coin is tossed. If heads appears, a spinner that can land on any number from 1 to 4 is spun. If tails appears, a second coin is tossed instead of spinning the spinner. What are the possible outcomes?
A.
H1 H2 H3 H4
B.
H1 H2 H3
C.
H1 H2 H3 H4 TH TT
D.
HH HT
The possible outcomes can be found by listing all the possible outcomes of the coin toss and the spinner (if heads is obtained) or the second coin toss (if tails are obtained).
The possible outcomes are:
A. H1, H2, H3, H4 (if heads appears and the spinner is spun)
B. H1, H2, H3 (if tails appear and the coin is tossed again)
C. H1, H2, H3, H4, TH (if tails appears and the coin is tossed instead of spinning the spinner)
D. HH, HT (if the first toss results in heads and the second toss results in heads or tails)
Therefore, the correct answer is option C: H1, H2, H3, H4, TH, and TT.
If the spinner is spun twice, what is the probability that the spinner will stop on a consonant and then again on a consonant?
The spinner is a circle divided into 6 equal sections. The sections are labeled L U Z O E and I.
A.
two-ninths
B.
start fraction 1 over 3 end fraction
C.
start fraction 1 over 6 end fraction
D.
one-ninth
The spinner has 6 sections, out of which 2 (L and Z) are consonants. That means the probability of the spinner stopping on a consonant on the first spin is 2/6 or 1/3. Since the section does not get replaced before the second spin, the total number of sections reduces to 5, with 1 less consonant (L or Z) and 4 vowels (U, O, E, and I). Therefore, the probability of the spinner stopping on a consonant on the second spin is 2/5.
To find the probability that the spinner will stop on a consonant and then again on a consonant, we need to multiply the probability of the first spin stopping on a consonant with the probability of the second spin stopping on a consonant.
So, the probability of the spinner stopping on a consonant and then again on a consonant is:
(1/3) * (2/5) = 2/15
Therefore, the answer is A: two-ninths.
In how many different ways can the following positions be chosen from the 18-member math team?
president, vice president, and treasurer
A.
4,080
B.
306
C.
5,814
D.
4,896
The president position can be filled in 18 ways, as any of the 18 members can be chosen. After the president is chosen, the vice-president position can be filled in 17 ways, as any of the remaining 17 members can be chosen. Finally, the treasurer position can be filled in 16 ways, as any of the remaining 16 members can be chosen.
Therefore, the total number of ways that the three positions can be filled is the product of the number of ways to fill each position:
18 * 17 * 16 = 4,896
Therefore, the answer is D: 4,896.
In Mr. Tarkiainen’s history class, each student chooses two of the topics listed in the chart above for their papers. How many different pairs can Juliana choose?
A chart with 2 columns is shown.
The first column is titled Year. The second column is titled Invention.
Row 1: 1807 Steamboat
Row 2: 1825 Photography
Row 3: 1837 Telegraph
Row 4: 1846 Sewing Machine
Row 5: 1869 Hydroelectric power
Row 6: 1876 Telephone
Row 7: 1879 light bulb
Row 8: 1888 Rabies Vaccine
Row 9: 1890 Diesel Engine
Row 10: 1897 Airplane
A.
90
B.
10
C.
45
D.
30
Juliana can choose 2 topics from the 10 topics listed in the chart above. The order in which she chooses the topics does not matter, since the question is only asking for the number of different pairs she can choose, not the order in which she chooses them.
The total number of ways that Juliana can choose 2 topics from 10 topics is:
(10 choose 2) = (10! / (2! * (10-2)!)) = (10 * 9) / 2 = 45
Therefore, Juliana can choose 45 different pairs of topics.
So, the answer is C: 45.
A multiple choice test has 5 questions each with 5 possible answers. Find the probability of answering all the questions correctly by guessing randomly.
A.
one over three-thousand-one-hundred-twenty-five
B.
one over twenty-five
C.
one over-six-hundred-twenty-five
D.
one over one-hundred-twenty-five
The probability of answering one question correctly by random guessing is 1/5, since there are 5 possible answers and only one of them is correct.
Since there are 5 questions, the probability of answering all 5 questions correctly by random guessing is:
P(all 5 correct) = (1/5)^5 = 1/3125
Therefore, the answer is A: one over three-thousand-one-hundred-twenty-five.
Eva wants to know the probability of tossing “heads” at least three times out of five. She used the random number table to simulate flipping a coin. She assigned odd digits to represent “heads” and even digits, including zero, to represent “tails.”
32766 53855 34591 27732
47406 31022 25144 72662
03087 35521 26658 81704
56212 72345 44019 65311
Use the table to find P(“heads” at least 3 out of 5 times).
To use the random number table to find the probability of tossing "heads" at least three times out of five, we can divide each of the 4-digit numbers into pairs and check each pair to see if it represents "heads" (odd) or "tails" (even). We can do this for each of the 20 digits in the table that represent the coin tosses.
For example, for the first number in the table (32766), we get the pairs 32, 76, and 6. The first pair represents "heads," the second pair represents "tails," and the third pair represents "tails." Since there are 2 "heads" out of 5 tosses, this outcome does not meet the requirement of getting at least 3 "heads."
We can repeat this process for each of the 20 numbers in the table and count the number of outcomes that meet the requirement of at least 3 "heads" out of 5 tosses.
Going through each of the 20 rows in the table, the outcomes that have at least 3 "heads" are:
- 53855: 3 "heads"
- 34591: 3 "heads"
- 72662: 3 "heads"
- 35521: 3 "heads"
- 81704: 3 "heads"
- 56212: 3 "heads"
- 72345: 3 "heads"
- 65311: 3 "heads"
- 32766: 4 "heads"
- 47406: 4 "heads"
- 31022: 4 "heads"
- 25144: 4 "heads"
Therefore, out of the 20 possible outcomes, 12 of them have at least 3 "heads." So, the empirical probability of getting at least 3 "heads" out of 5 tosses based on the random number table is approximately:
P(at least 3 "heads") = 12/20 = 0.6
Therefore, the answer is not one of the multiple choices given.