Answer:
At a certain pizza parlor,36 % of the customers order a pizza containing onions,35 % of the customers order a pizza containing sausage, and 66% order a pizza containing onions or sausage (or both). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage.
Step-by-step explanation:
Hello!
You have the following possible pizza orders:
Onion ⇒ P(on)= 0.36
Sausage ⇒ P(sa)= 0.35
Onions and Sausages ⇒ P(on∪sa)= 0.66
The events "onion" and "sausage" are not mutually exclusive, since you can order a pizza with both toppings.
If two events are not mutually exclusive, you know that:
P(A∪B)= P(A)+P(B)-P(A∩B)
Using the given information you can use that property to calculate the probability of a customer ordering a pizza with onions and sausage:
P(on∪sa)= P(on)+P(sa)-P(on∩sa)
P(on∪sa)+P(on∩sa)= P(on)+P(sa)
P(on∩sa)= P(on)+P(sa)-P(on∪sa)
P(on∩sa)= 0.36+0.35-0.66= 0.05
I hope it helps!
If I’m not mistaken I think this is what you’re asking for?
Answer:
The answer is 31 pennies
Step-by-step explanation:
we have to find a number that gives a remainder of 1 when divided by both 2, 3 and 5.
The easiest way to do this is to list the factors of 5, add 1 to it, and test them until we find the one with a remainder of 1 by all three divisors. This is done as follows:
1) factor: 5 (+1) = 6
6 ÷ 2 = 3 remainder 0
6 ÷ 3 = 2 remainder 0
since there is a remainder of 0, when divided by 3, 6 is wrong
2) factor; 10 (+1) = 11
11 ÷ 3 = 3 remainder 2 ( 11 is not the correct answer
3) factor : 15 (+1) = 16
16 ÷ 2 = 8 remainder 0 ( 16 is not the correct answer
4) factor : 20(+1) = 21
21 ÷ 3 = 7 remainder 0 ( 21 is wrong)
5) factor : 25(+1) = 26
26 ÷ 2 = 13 remainder 0 ( 26 is wrong)
6) factor : 30(+1) = 31
31 ÷ 2 = 15 Remainder 1
31 ÷ 3 = 10 Remainder 1
31 ÷ 5 = 6 Remainder 1
since all three divisors give a remainder of 1, the correct answer is 31
Therefore you have 31 pwnnies
Answer:
first do:
60mph * 4 hours= 240mi
Let D = the distance the passenger train has to
travel to catch the freight train
Start a stopwatch when the passenger train leaves
Let t = the time on the stopwatch when they meet
Equation for freight train:
d-240=60t
Equation for passenger train:
d=100t
all together:
100t-240=60t
we can do 100t-60t because they both have (t)= 40t
40t=240 *now divide 240 by 40= 6
THEN do:
D= 100mph x 6= 600mi
They will meet 600 mi from the station
to check:
d-240=60t x 6
600-240= 360
60 x 6= 360
360=360
The system of equations has one solution
