Answer:
Step-by-step explanation:
The probability of getting one even and one odd is the sum of probability of getting even and of getting odd and even. Since there are 3 even or odd on each die out of six numbers you have
P(EO)(3/6)(3/6)=1/4 and P(OE)=(3/6)(3/6)=1/4 so 1/4+1/4= 1/2
or you can find the probability of getting two odds or two evens and subtract that from 1
P(EE)=(3/6)(3/6)=1/4, P(OO)=(3/6)(3/6)=1/4
P(EO)=1-1/4-1/4=1/2
1/2 = 0.5
Proof:
1/4 + 1/4 = 2/4 = 1/2 = 0.25 + 0.25 = 0.50
Part A:
[$25-( $0.50×22)]÷$2=n
Part B:
[$25-($0.50×22)]÷$2=n
[$25-($11)]÷$2=n
[$14]÷$2=n
7=n
Part C: Brandon purchased 7 meals last month.
Answer:
yes, the meal costs 25.60
Step-by-step explanation:
I used a slightly different method but B (Yes. They have enough to pay their bill: 21 dollars times 1.06 = 22 dollars and 26 cents and 22 dollars and 26 cents times 1.15 = 25 dollars and 60 cents.) is correct.
21*0.06=1.26
1.26+21=22.26
22.26*0.15=3.34
22.26+3.34=25.60
Answer:
Step-by-step explanation:
The probability of asking for the cherries flambe and dying, P(A) = 0.5 × 0.6 = 0.3
The probability of asking for the chocolate mousse and dying, P(A) = 0.4 × 0.9 = 0.36
The probability of skipping dessert altogether = 0.1 × 0 = 0
The probability that the rich uncle dies, P(D) is
P(D) = 0.3 + 0.36 + 0 = 0.66
To determine the prime suspect, we would apply the concept of conditional probability. It is the probability of an event, A occurring given that event B has occurred. We would compare probabilities of the events A and B occurring given that event D also occurred
Therefore,
P(A|D) = 0.3/0.66 = 0.45
P(B|D) = 0.36/0.66 = 0.55
Therefore, the prime suspect is Margo because his probability is higher
