The outlier is 51.
The mean is (12+51+21+18+24+17+12+25)/8 = 23.5.
The mean without the outlier is (12+21+18+24+17+12+25)/7 = 18.14 (rounded to two decimal places).
What is the outlier? 12, 51, 21, 18, 24, 17, 12, 25
What is the mean? (Add all 8 numbers together and divide by 8) and What is the mean without the outlier? (take out the outlier, add up the remaining 7 numbers and divide by 7)
9 answers
Use the table to answer the question.
Number of White Balls Selected Number of Red Balls Selected
Bowl A 45 55
Bowl B 60 40
Bowls A and B contain a number of white and red balls. Clark repeatedly selected a ball from both bowls and recorded the results in a table. If there are 500 balls in Bowl B, what is the estimated difference in the expected number of white and red balls in Bowl B?
Responses
50
50
200
200
300
300
Number of White Balls Selected Number of Red Balls Selected
Bowl A 45 55
Bowl B 60 40
Bowls A and B contain a number of white and red balls. Clark repeatedly selected a ball from both bowls and recorded the results in a table. If there are 500 balls in Bowl B, what is the estimated difference in the expected number of white and red balls in Bowl B?
Responses
50
50
200
200
300
300
The estimated difference in the expected number of white and red balls in Bowl B is (60/100 - 40/100) x 500 = 20 x 5 = 100.
So the answer is 100.
So the answer is 100.
What is the probability of rolling an odd number on the first roll of a six-sided cube and rolling an even number on the second roll?
Responses
14
Start Fraction 1 over 4 End Fraction
2
2
12
, Start Fraction 1 over 2 End Fraction
1
Responses
14
Start Fraction 1 over 4 End Fraction
2
2
12
, Start Fraction 1 over 2 End Fraction
1
The probability of rolling an odd number on the first roll of a six-sided cube is 3/6 or 1/2 (since there are three odd numbers and six total numbers). The probability of rolling an even number on the second roll, given that an odd number was rolled on the first roll, is also 3/6 or 1/2 (since there are three even numbers out of the remaining six possible outcomes).
Therefore, the overall probability of these two events happening together is:
1/2 x 1/2 = 1/4
So the answer is Start Fraction 1 over 4 End Fraction.
Therefore, the overall probability of these two events happening together is:
1/2 x 1/2 = 1/4
So the answer is Start Fraction 1 over 4 End Fraction.
A big flower vase has 5 begonias, 6 carnations, and 7 tulips. Two flowers are randomly selected without replacement. What is the probability of selecting two carnations? Express the answer in the simplest form of fraction.
1$$
Correct answers:
1551
1$$
Correct answers:
1551
The probability of selecting one carnation from the vase is 6/(5+6+7) = 6/18 = 1/3.
After one carnation is selected, there are only 5 carnations left out of 16 total flowers remaining. So the probability of selecting a second carnation from the vase, given that a carnation was already selected on the first draw, is 5/16.
The probability of selecting two carnations, without replacement, is the product of these two probabilities:
(1/3) x (5/16) = 5/48
So the answer is 5/48, which is already expressed in simplest form.
After one carnation is selected, there are only 5 carnations left out of 16 total flowers remaining. So the probability of selecting a second carnation from the vase, given that a carnation was already selected on the first draw, is 5/16.
The probability of selecting two carnations, without replacement, is the product of these two probabilities:
(1/3) x (5/16) = 5/48
So the answer is 5/48, which is already expressed in simplest form.
A big flower vase has 5 begonias, 6 carnations, and 7 tulips. Two flowers are randomly selected without replacement. What is the probability of selecting two carnations? Express the answer in the simplest form of fraction.
1$$
Correct answers:
15/51
1$$
Correct answers:
15/51
The probability of selecting one carnation from the vase is 6/(5+6+7) = 6/18 = 1/3.
After one carnation is selected, there are only 5 carnations left out of 16 total flowers remaining. So the probability of selecting a second carnation from the vase, given that a carnation was already selected on the first draw, is 5/16.
The probability of selecting two carnations, without replacement, is the product of these two probabilities:
(1/3) x (5/16) = 5/48
So the answer is 5/48, which is not equivalent to 15/51.
After one carnation is selected, there are only 5 carnations left out of 16 total flowers remaining. So the probability of selecting a second carnation from the vase, given that a carnation was already selected on the first draw, is 5/16.
The probability of selecting two carnations, without replacement, is the product of these two probabilities:
(1/3) x (5/16) = 5/48
So the answer is 5/48, which is not equivalent to 15/51.