Answer:
Part 1 , not significant
Part II, significant
Step-by-step explanation:
Given that a heterozygous white-fruited squash plant is crossed with a yellow-fruited plant, yielding 200 seeds. of these seeds, 110 produce white-fruited plants while only 90 produce yellow-fruited plants

(Two tailed chi square test)
We assume H0 to be true and find out expected
If H0 is true expected would be 100 white and 100 yellow
Chi square = 
df = 1
p value = 0.152799
Since p value > 0.05 at 5% level we accept that the colours are equally likely
2)
Here observed are 1100 and 900
Expected 1000 & 1000
df = 1
Chi square = 
p value <0.0001
These results are here statistically significant.
Team A) 45 people
Team B) 55 people
A)There are two ways to solve this problem, finding the number of combinations possible for Team B, or the number of combinations possible for Team A.
Team A
It's a given that 20 mathematicians are on team A, which leavs the other 25 people for team A to be chosen from a pool of 80 (100- 20 mathletes)
80-C-25 = 80! / (25!/(80-25)!) =<span>363,413,731,121,503,794,368
</span>or 3.63 x 10^20
Solving using Team B
Same concept, but choosing 55 from a pool of 80 (mathletes excluded)
80-C-25 = 80! / (55!(80-55!) = 363,413,731,121,503,794,368
or 3.63 x 10^20
As you can, we get the same answer for both.
B)
If none of the mathematicians are on team A, then we exclude the 20 and choose 45:
80-C-45 = 80! / (45!(80-45)!) = <span>5,790,061,984,745,3606,481,440
or 5.79 x 10^22
Note that, if you solve from the perspective of Team B (80-C-35), you get the same answer</span>
Answer: That would be 1 case + 112 pack + 16 pack + 5 singles
Step-by-step explanation:
It would be A or B, as an integer can be anything and it’s a while number.