Sebastian solved the radical equation y + 1 = but did not check his solution. (y + 1)2 = y2 + 2y + 1 = –2y – 3 y2 + 4y + 4 = 0 (
1 answer:
You might be interested in
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.