Your answer will be 7.50. What I did was take 1.25 and times it by 6.
The axis of symmetry is 1.
To find this, do -b/2a, putting -8 in the place of b and -4 in the place of a. Proceed to solve to get 1.
The equation provided is vertex form, so to find the needed components, you need to solve for the vertex. To do so, implement 1 in as x and solve for the vertex.
The vertex is 1, -16.
(y - 16) = -4(x - 1)^2
Although it is not exactly vertex form, the only difference is that instead of it being +k at the end, it is subtracted with the y.
Hope this helps!
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 ![\lambda = \frac{10.2}{200} = 0.051](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B10.2%7D%7B200%7D%20%3D%200.051)
a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar
![\frac{1}{4} \times 200 = 50](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D%20%5Ctimes%20200%20%3D%2050)
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.
![P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004](https://tex.z-dn.net/?f=P%28X%3E10%29%3D%20e%5E%7B-%5Clambda%20%5Ctimes%20x%7D%3D%20e%5E%7B-0.051%20%5Ctimes%2010%7D%3D%20e%5E%7B-0.51%7D%3D0.6004)
c)The expected number of grams to be eaten before encountering the first fragments :
s