HELP PLS!!! IMPORTANT!
1 answer:
Answer:
#3: The number of pets that is the mode for the class is 1 because the people who have 1 pet appears the most.
explanation: the largest amount of people are the people who have 1 pet.
#4: The percent of students who have at least 3 pets is 25%.
explanation: 6 divided by 24 is 0.25 and and that converted to a percentage is 25%
#5: The median number of pets that students have is 1 pet.
explanation: the middle number of all the people in Yolanda's class is 1.
#6: The range of the number of pets is 9.
explanation: range is the largest number minus the smallest number
9 - 0 = 9
#7: #7 will be attached with this answer.
#8: There are 22 students in the class.
explanation: the added up number of students is 22.
#9: The mode value is 3 students.
explanation: the mode is the number that appears the most. 3 is the number that appears the most.
#10: The median number of concerts that students have attended is 2 concerts.
explanation: the middle number of all the numbers is 2.
#11: The concerts that the students in Mr. Ritter's class have attended in all is 72.
explanation: all the number of concerts attended added up is 72.
#12: The average (mean) number of concerts the students in Mr. Ritter's class has attended is 3 concerts.
explanation: all the concerts divided by the number of students were are is the mean.
72 / 22 is 3.27.
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Answer: The calculations are below.
Step-by-step explanation: The calculations are as follows:
(3) Given that for two events A and B,
Since A and B are disjoint, so
From the law of probability, we have
Thus, the correct option is (E) 0.50.
(4) Given that a fair six-sided die is rolled. We are to find the probability that an odd number is rolled.
Let, 'A' be event of rolling an odd number. So,
A = {1, 3, 5} ⇒ n(A) = 3.
Let 'S' be the sample space for the experiment. So,
S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6.
Therefore, the probability of rolling an odd number is given by
Thus, the correct option is (D)
(5) Given that there are 3 red marbles, 4 white marbles, and 1 green marble in a bag and marbles are drawn without replacement.
So, the probability that 3 marbles can be drawn without drawing the green marble is given by
Thus, the correct option is (A) 0.625.
(6) The sample space for rolling a six-sided die is {1, 2, 3, 4, 5, 6}
and the sample space for tossing a coin is {H, T}.
Therefore, the sample space for rolling a six-sided die and tossing a coin will be
S = {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}.
Thus, the correct option is {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}.
(7) Given that for two events A and B,
Since A and B are independent but not necessarily disjoint, so
From the law of probability, we have
Thus, the correct option is (C) 0.44.
(8) Given that three coins are tossed.
The sample space for each of them is S = {1, 2, 3, 4, 5, 6}.
The equally likely outcomes when three coins are tossed together are
{(1, 1 , 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6)}.
So, the total number of equally likely outcomes = 6.
Thus, the correct option is (C) 6.
Hence all the questions are answered.