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Answer:
a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55
b)0.6004
c)19.607
Step-by-step explanation:
Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2
X ~ Poisson(A) where
a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar
50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55
So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55
b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment
Let X denotes the number of grams to be eaten before another fragment is detected.
c)The expected number of grams to be eaten before encountering the first fragments :
s
1) The length of the shadow the tree cast and the shadow Bill cast needs to be added together which makes 34 ft
2) After setting up Bill's Height and Shadow over each other 6ft/9ft.. Set that equal to another proportion...
3) The next proportion is x (the height of the tree) and the length (along the bottom) X/34
4 Solveeee :) 6/9 multiplied by 34 is 22.6- 22 ⅔ft as a fraction
andddd you're done that's the answer
Hope this helpppsss :)
2) After setting up Bill's Height and Shadow over each other 6ft/9ft.. Set that equal to another proportion...
3) The next proportion is x (the height of the tree) and the length (along the bottom) X/34
4 Solveeee :) 6/9 multiplied by 34 is 22.6- 22 ⅔ft as a fraction
andddd you're done that's the answer
Hope this helpppsss :)