I think the ratio is for for every 4 packs it cost 10 dollars so in all it would cost 30 dollars
The answer I got was A.)$56.
Hope this helps!<3
Comment if you need help on how I got that answer!
The three numbers are 2, 5, and 11.
The sum of 3 numbers is 18. 2 + 5 + 11 = 18
Since the problem states that every number is a prime number, then the number must be a natural number greater than 1 that has no positive divisors other than 1 and itself.
2 / 1 = 2 OR 2 / 2 = 1 These numbers are prime numbers because their divisors
5 / 1 = 5 OR 5 / 5 = 1 are 1 and itself.
11/1 = 11 OR 11/11 = 1
Answer:
I think [(-4)^4]^5, but I'm not completely sure
Step-by-step explanation:
1.0995116e+12 greater than -4.7223665e+21
[(-4)^4]^5=1.0995116e+12
-[(4^12)^3=-4.7223665e+21
Answer:
In a certain Algebra 2 class of 30 students, 22 of them play basketball and 18 of them play baseball. There are 3 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?
I know how to calculate the probability of students play both basketball and baseball which is 1330 because 22+18+3=43 and 43−30 will give you the number of students plays both sports.
But how would you find the probability using the formula P(A∩B)=P(A)×p(B)?
Thank you for all of the help.
That formula only works if events A (play basketball) and B (play baseball) are independent, but they are not in this case, since out of the 18 players that play baseball, 13 play basketball, and hence P(A|B)=1318<2230=P(A) (in other words: one who plays basketball is less likely to play basketball as well in comparison to someone who does not play baseball, i.e. playing baseball and playing basketball are negatively (or inversely) correlated)
So: the two events are not independent, and so that formula doesn't work.
Fortunately, a formula that does work (always!) is:
P(A∪B)=P(A)+P(B)−P(A∩B)
Hence:
P(A∩B)=P(A)+P(B)−P(A∪B)=2230+1830−2730=1330
