Answer:
8 home games and 10 away games
Step-by-step explanation:
Total home goals
= 8 +5 +9 +8 = 30
Number of home games
= 30/3.75 = 8
Total away game goals
= 7 + 8 + 4 + 5 = 24
Number of away games
= 24/2.4 = 10
Easy Peasy :) Hope this helped! <3 brainliest please :)
Let X be the number of boys in n selected births. Let p be the probability of getting baby boy on selected birth.
Here n=10. Also the male and female births are equally likely it means chance of baby boy or girl is 1/2
P(Boy) = P(girl) =0.5
p =0.5
From given information we have n =10 fixed number of trials, p is probability of success which is constant for each trial . And each trial is independent of each other.
So X follows Binomial distribution with n=10 and p=0.5
The probability function of Binomial distribution for k number of success, x=k is given as
P(X=k) = 
We have to find probability of getting 8 boys in n=10 births
P(X=8) = 
= 45 * 0.0039 * 0.25
P(X = 8) = 0.0438
The probability of getting exactly 8 boys in selected 10 births is 0.044
<span>the discriminant formula is D = b^2 - 4ac
(idk how to put graphs or anything but yeh)
</span>
Answer:
The answer of the following question is m = \frac{C - b - bt}{r + rt}.
Solution:
C = (b + rm)(1 + t),
C = b + rm + bt + rmt
C = b + bt + rm + rmt
C - b - bt = m (r + rt)
\frac{C - b - bt}{r + rt} = m
t\neq -1,
r\neq 0