x= | 1 < x < 3
x + 2 < 5 and x - 7 > -6
x < 5 - 2 and x > -6 + 7
I don't know if I am 100% right.
Answer:
d=~7.6
Step-by-step explanation:
7^2+3^2=d^2
49+9=d^2
58=d^2
Square root of 58=~7.6
d=~7.6
The possible digits are:
5, 6, 7, 8 and
9. Let's mark the case when the locker code begins with a prime number as
A and the case when <span>the locker code is an odd number as
B. We have
5 different digits in total,
2 of which are prime (
5 and
7).
First propability:
</span>

<span>
By knowing that digits don't repeat we can say that code is an odd number in case it ends with
5, 7 or
9 (three of five digits).
Second probability:
</span>
Answer:
First question: -0.577 cannot be a probability.
Second question: The number must be between 0 and 1, inclusive (D)
Third question: -0.577 cannot be a probability.
Step-by-step explanation:
A probability is a real number that is in the interval [0, 1]. Therefore, any number outside this range cannot be considered a probability. Under this premise:
First question: -0.577 cannot be a probability.
Second question: The number must be between 0 and 1, inclusive (D)
Third question: -0.577 cannot be a probability.
Answer:
P(B|A)=0.25 , P(A|B) =0.5
Step-by-step explanation:
The question provides the following data:
P(A)= 0.8
P(B)= 0.4
P(A∩B) = 0.2
Since the question does not mention which of the conditional probabilities need to be found out, I will show the working to calculate both of them.
To calculate the probability that event B will occur given that A has already occurred (P(B|A) is read as the probability of event B given A) can be calculated as:
P(B|A) = P(A∩B)/P(A)
= (0.2) / (0.8)
P(B|A)=0.25
To calculate the probability that event A will occur given that B has already occurred (P(A|B) is read as the probability of event A given B) can be calculated as:
P(A|B) = P(A∩B)/P(B)
= (0.2)/(0.4)
P(A|B) =0.5