Answer:
0.1505 = 15.05% probability that the hockey team wins 6 games in November
Step-by-step explanation:
For each game, there are only two possible outcomes. Either the team wins, or it does not. The probability of winning a game is independent of winning other games. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which 
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
The probability that a certain hockey team will win any given game is 0.3723
So 
12 games in November
So 
What is the probability that the hockey team wins 6 games in November?
This is 
0.1505 = 15.05% probability that the hockey team wins 6 games in November
Answer:
1a. p0= 0.714
1b The result is not reasonably close to the value of 3/4 that was expected
Step-by-step explanation:
1a.Since One sample of offspring contained 376 green peas and 150 yellow peas therefore the probability of getting an offspring pea that is green will be:
Green pea/(Green pea+ Yellow pea)
p0= 376 /(376+150)
p0=376/526
Probability of getting Green pea = 0.714
1b.The result is not reasonably close to the value of 3/4 that was expected.
Answer:
28.07 and rounded to the nearest hundredth would be 28.10
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
the "-" sign can't be true, because that would have turned the whole graph of x² upside-down. but it is not.
so, we need a positive sign for the x term.
and the whole graph is shifted to the right. so the functional value of x of F(x) is now the same that happened "earlier" (for smaller x) with G(x). so, the argument "x" is now reduced (by 3 units, as the graph is shifted right by 3 units) in the calculation compared to the original G(x) calculation.
therefore, the ...(x-3)... expression is right.
so, positive sign and "(x-3)" is only in C.
Bart and David are correct.
Bart
6x(1+2/3^3)
6x * 1 = 6x
6x *2/3^3 = 4x^3
6x * 4x^3
David
x(6+4x^2)
x*6 = 6x
x*4x^2 = 4x^3
6x+4x^3
