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Eduardwww [97]
3 years ago
9

Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2

00-gram chocolate bar being 10.2. 15. What is the expected number of insect fragments in 1/4 of a 200-gram chocolate bar? 16. What is the probability that you have to eat more than 10 grams of chocolate bar before nding your rst fragment? 17. What is the expected number of grams to be eaten before encountering the rst fragment?
Mathematics
1 answer:
leonid [27]3 years ago
7 0

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

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

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

c)The expected number of grams to be eaten before encountering the first fragments :

E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

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IrinaK [193]

Answer:

11.

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13.

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15.

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7 0
2 years ago
If side A is twice as long as B and C is 25 using the Pythagorean Theorem,What are the lengths of side A and B? Round to the nea
777dan777 [17]

Answer:

<u>The lengths of side A is 22.4 and B is 11.9</u>.

Step-by-step explanation:

Given:

If side A is twice as long as B and C is 25 using the Pythagorean Theorem.

Now, to find the lengths of side A and B.

Let the side B be x.

So, the side A be 2x.

Side C = 25.

Now, to solve by using Pythagorean Theorem:

A² + B² = C²

(2x)^2+(x)^2=(25)^2

4x^2+x^2=625

5x^2=625

<em>Dividing both sides by 5 we get:</em>

x^2=125

<em>Using square root on both sides we get:</em>

x=11.18.

<u>B rounding to the nearest tenth =  11.9.</u>

Now, to get A by substituting the value of x:

2x\\=2\times 11.18\\=22.36.

<u>A rounding to the nearest tenth =  22.4.</u>

Therefore, the lengths of side A is 22.4 and B is 11.9.

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3 years ago
Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shap
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Answer:

0.25 feet per minute

Step-by-step explanation:

Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min. Since we are told that the shape formed is a cone, the rate of change of the volume of the cone.

\dfrac{dV}{dt}=20$ ft^3/min

\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h

Since base diameter = Height of the Cone

Radius of the Cone = h/2

Therefore,

\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}

\text{Rate of Change of the Volume}, \dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}

Therefore: \dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=20

We want to determine how fast is the height of the pile is increasing when the pile is 10 feet high.

h=10$ feet$\\\\\dfrac{3\pi *10^2}{12}\dfrac{dh}{dt}=20\\25\pi \dfrac{dh}{dt}=20\\ \dfrac{dh}{dt}= \dfrac{20}{25\pi}\\ \dfrac{dh}{dt}=0.25$ feet per minute (to two decimal places.)

When the pile is 10 feet high, the height of the pile is increasing at a rate of 0.25 feet per minute

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kenny6666 [7]
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marta [7]

Answer:

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48 = 5x + 2(15 - x)

48 = 5x + 30 - 2x

48 = 3x + 30

18 = 3x

x = 6

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