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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
XY = 17.
Because YW is a perpendicular bisector, we can say that TW and WZ are both equal to 3. It tells us that XZ is 12, so that means that XW must be 12+3 which equals 15.
It also tells us that YW is 8. So we can use the Pythagorean Theorem to find the hypotenuse of Triangle XWY. Thus,
To solve this equation, square and add both of the terms on the right like this:
And then take the square root of both sides. Your final answer should be:
XY = 17.
Because YW is a perpendicular bisector, we can say that TW and WZ are both equal to 3. It tells us that XZ is 12, so that means that XW must be 12+3 which equals 15.
It also tells us that YW is 8. So we can use the Pythagorean Theorem to find the hypotenuse of Triangle XWY. Thus,
To solve this equation, square and add both of the terms on the right like this:
And then take the square root of both sides. Your final answer should be:
XY = 17.