the snack preferences of 50 shoppers at the mall. what is the probability that one shopper selected at random from 50 surveyed perfered the potato chips or pretzels
Without any additional information, we cannot determine the exact probabilities. We would need to know the number of shoppers who prefer potato chips, the number of shoppers who prefer pretzels, and if there are any shoppers who prefer both. Then we could use that information to calculate the probabilities.
12 for potato chips and 14 for pretzels
In this case, out of the 50 shoppers surveyed, 12 preferred potato chips and 14 preferred pretzels.
So the probability that a shopper selected at random from the 50 surveyed prefers either potato chips or pretzels would be:
P(potato chips or pretzels) = P(potato chips) + P(pretzels) - P(both)
= 12/50 + 14/50 - 0
= 26/50
= 0.52
Therefore, the probability that one shopper selected at random from 50 surveyed preferred the potato chips or pretzels is 0.52 or 52%.
in a fraction
The probability can also be expressed as a fraction:
P(potato chips or pretzels) = 26/50
which simplifies to:
P(potato chips or pretzels) = 13/25
a standard number cube has numbers 1 through 6 what is the probability of rolling a number greater than 4
Out of the six possible outcomes when rolling a standard number cube (1, 2, 3, 4, 5, and 6), three numbers are greater than 4: 5 and 6.
So the probability of rolling a number greater than 4 would be:
P(number greater than 4) = number of favorable outcomes / total number of outcomes
= 2 / 6
= 1 / 3
Therefore, the probability of rolling a number greater than 4 is 1/3 or approximately 0.333.
a number cube is rolled 360 times and the results as the following 41 ones, 54 twos, 62 threes, 74 fours, 35 fives, 95 sixes. what is the experimental probability of rolling a two or a three
To find the experimental probability of rolling a two or a three, we need to add the frequencies of rolling a two and a three, since these are the favorable outcomes in this case.
Frequency of rolling a two = 54
Frequency of rolling a three = 62
So the total frequency of rolling a two or a three would be:
54 + 62 = 116
The total number of rolls was 360, so the experimental probability of rolling a two or a three would be:
experimental probability = frequency of favorable outcomes / total number of outcomes
= 116/360
= 0.3222 (rounded to four decimal places)
Therefore, the experimental probability of rolling a two or a three is approximately 0.3222 or 32.22% (rounded to the nearest hundredth).
From a barrel of colored marbles you randomly select 5 blue, 2 yellow, 3 red, 4 green, and 2 purple marbles. Find the experimental probability of randomly selecting a marble that is not yellow
The total number of marbles in the barrel is: 5 + 2 + 3 + 4 + 2 = 16
The number of marbles that are not yellow is: 5 + 3 + 4 + 2 = 14
So the experimental probability of randomly selecting a marble that is not yellow would be:
experimental probability = number of favorable outcomes / total number of outcomes
= 14/16
= 0.875 (rounded to three decimal places)
Therefore, the experimental probability of randomly selecting a marble that is not yellow is approximately 0.875 or 87.5%.
in a fraction
The experimental probability can also be expressed as a fraction:
experimental probability = number of favorable outcomes / total number of outcomes
= 14/16
which simplifies to:
experimental probability = 7/8
Therefore, the experimental probability of randomly selecting a marble that is not yellow is 7/8.
a survey showed that 56% of car owners prefer four door cars, 31% two door cars and 13% preference. you ask 300 people how many do you think will prefer the four door cars
If 56% of car owners prefer four door cars, we can expect that approximately:
0.56 x 300 = 168
people out of 300 would prefer four door cars.
Similarly, if 31% prefer two door cars, we can expect that approximately:
0.31 x 300 = 93
people out of 300 would prefer two door cars.
And if 13% have no preference, we can expect that approximately:
0.13 x 300 = 39
people out of 300 would have no preference.
Note that these are expected values based on the given percentages and the total sample size of 300. The actual results may differ due to sampling error or other factors.
Tony works in a factory that produces 1,000 each day when 50 were sampled it was found that 7 were defective. estimate computer are made each day
If 7 out of 50 computers sampled were found to be defective, we can use this proportion to estimate the number of defective computers produced each day:
proportion of defective computers = number of defective computers / total number of computers
We can assume that this proportion is the same for the daily production and the sampled computers. Therefore:
7/50 = x/1000
where x is the estimated number of defective computers made each day.
To solve for x, we can cross-multiply:
50x = 7 * 1000
x = (7 * 1000) / 50
x = 140
Therefore, we can estimate that 140 computers produced each day are defective.
what are 55 good songs from Kanye west
As an AI language model, I cannot provide with that list since this is subjective. However, here are some popular and critically acclaimed Kanye West songs:
1. Jesus Walks
2. All Falls Down
3. Gold Digger
4. Stronger
5. Good Life
6. Heartless
7. Power
8. Runaway
9. All of the Lights
10. Monster
11. Ultralight Beam
12. Famous
13. Father Stretch My Hands Pt. 1
14. Waves
15. No More Parties in LA
16. Saint Pablo
17. Black Skinhead
18. New Slaves
19. Blood on the Leaves
20. Bound 2
21. We Major
22. Street Lights
23. Flashing Lights
24. Love Lockdown
25. Paranoid
26. Welcome to Heartbreak
27. Homecoming
28. Champion
29. Touch the Sky
30. Diamonds from Sierra Leone
31. Heard 'Em Say
32. Hey Mama
33. Roses
34. Gone
35. Good Morning
36. The Glory
37. Can't Tell Me Nothing
38. Everything I Am
39. The People Upstairs
40. So Appalled
41. Devil In A New Dress
42. See Me Now
43. Real Friends
44. Wolves
45. Fade
46. 30 Hours
47. Feedback
48. Father Stretch My Hands Pt. 2
49. Famous Pt. 2
50. Heartless Pt. 2
51. as in Paris
52. Watch The Throne
53. Otis
54. Lift Off
55. H.A.M.
the spinner is divided into equal parts use a tree diagram use a tree diagram to find the probability that the spinner is spun twice
I'm sorry, I'm not sure I understand the question. Can you please provide more information?
What are the parts the spinner is divided into? How many parts are there? What are we trying to determine the probability of?
3
If the spinner is divided into three equal parts, we can use a tree diagram to show the possible outcomes of spinning the spinner twice.
The first spin can result in three possible outcomes: A, B, or C. Each of these outcomes can then be followed by another spin, resulting in a total of nine possible outcomes.
Here is a tree diagram that shows the possible outcomes:
```
|---- A ----|
| |
Start --| |---- A-A
| |
| |---- A-B
|
|---- B ----|
| |
| |---- B-A
| |
| |---- B-B
|
|---- C ----|
|
|---- C-A
|
|---- C-B
```
To find the probability of an event, we can sum the probabilities of all the outcomes that make up that event. For example, if we want to find the probability of spinning the spinner twice and getting the same letter both times, we can add up the probabilities of the outcomes A-A, B-B, and C-C:
P(same letter both times) = P(A-A) + P(B-B) + P(C-C)
To find the probability of each outcome, we need to know the probability of each letter appearing on the spinner. Since the spinner is divided into three equal parts, each letter has a probability of 1/3.
Therefore, the probability of getting the same letter both times would be:
P(same letter both times) = (1/3) * (1/3) + (1/3) * (1/3) + (1/3) * (1/3)
= 3/9
= 1/3
Therefore, the probability of spinning the spinner twice and getting the same letter both times is 1/3 or approximately 0.333.
an ice cream shop offers 5 different flavors of ice cream and 9 different toppings are possible for a single serving of ice cream with one topping
If the ice cream shop offers 5 different flavors of ice cream and 9 different toppings, the total number of possible combinations of ice cream and one topping would be:
5 (number of ice cream flavors) x 9 (number of possible toppings) = 45
Therefore, there are 45 possible single serving combinations of ice cream and one topping.
Note that if multiple toppings were allowed, the number of possible combinations would increase. For example, if customers could choose two toppings, the total number of possible combinations would be:
5 (number of ice cream flavors) x 9 (number of possible first toppings) x 9 (number of possible second toppings) = 405
Because there would be 9 possible second toppings for each first topping, resulting in a total of 9 x 9 = 81 possible combinations for each ice cream flavor. Then, since there are 5 ice cream flavors, we would multiply 81 x 5, which gives us 405.
which events are not independent
you toss three coins and get one head and one tail
you choose three different ice toppings for a sundae
you draw draw two colored pencils without replacement and get one red and one yellow
you pull a yellow marble from a bag of marbles return it and then pull a green marble
The events that are not independent are:
- You draw two colored pencils without replacement and get one red and one yellow. The probability of drawing a red pencil on the second draw depends on the outcome of the first draw. If you have already drawn a red pencil on the first draw, then the probability of drawing another red pencil on the second draw is zero. Therefore, the outcomes of the two draws are not independent.
- You pull a yellow marble from a bag of marbles, return it, and then pull a green marble. Since the yellow marble is returned to the bag before the second draw, each draw is independent, but the two draws as a whole are not independent. The probability of pulling a yellow marble on the second draw may be affected by the outcome of the first draw, because the yellow marble may or may not have been returned to the bag after the first draw.
if a coin is tossed and a number cube is rolled what is the probability that the coin lands an heads and the number cube land s on three
If a coin is tossed and a number cube is rolled, and we assume that the coin and number cube are fair and independent, we can find the probability of getting heads on the coin and rolling a three on the number cube by multiplying the probabilities of each event:
P(coin lands on heads and number cube lands on three) = P(coin lands on heads) * P(number cube lands on three)
Since the coin has two equally probable outcomes (heads or tails) and the number cube has six equally probable outcomes (one through six), we have:
P(coin lands on heads) = 1/2
P(number cube lands on three) = 1/6
So the probability of getting heads on the coin and rolling a three on the number cube would be:
P(coin lands on heads and number cube lands on three) = P(coin lands on heads) * P(number cube lands on three)
= (1/2) * (1/6)
= 1/12
Therefore, the probability of the coin landing on heads and the number cube landing on three is 1/12 or approximately 0.083.
on Friday the local diner serves 3 main entrees each with a choice of vegetable the entrees are beef, chicken, and fish. the veggies are spinach carrots and broccoli how many possible entrees and vegetable combos does the dinners serve list them
Since there are 3 main entrees and 3 vegetable choices, we can determine the total number of possible entree and vegetable combinations by multiplying the number of entrees by the number of vegetable choices:
3 (number of entrees) x 3 (number of vegetable choices) = 9
Therefore, there are 9 possible entree and vegetable combinations. They are:
1. Beef and spinach
2. Beef and carrots
3. Beef and broccoli
4. Chicken and spinach
5. Chicken and carrots
6. Chicken and broccoli
7. Fish and spinach
8. Fish and carrots
9. Fish and broccoli
So the diner serves 9 possible entree and vegetable combos.