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Artist 52 [7]
3 years ago
15

Nigel travels a total of 312 miles to school each day. He walks 14 mile to the bus stop. He travels 234 miles on the city bus. I

n miles, how far does the bus drop him off from the school?
Mathematics
1 answer:
Valentin [98]3 years ago
7 0
The bus drops Nigel off 64 miles away from the school.

Step by step Explanation:

An equation for this would be 312-14-234 (this was the equation I used. Since you would do left to write to solve the equation, you would do, 312-14 which equals 298. Then, the equation is left as 298-234, and when you solve that part to the equation, yoh would be getting 64. And that's how you would get the answer 64miles away from the school.
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Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie
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Answer:

\lambda \geq 6.63835

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that X \sim Poisson(\lambda)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

P(X\geq 2)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}

P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}

And replacing we have this:

P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)

And we want this probability that at least of 99%, so we can set upt the following inequality:

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99

And now we can solve for \lambda

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\lambda-ln(1+\lambda)+ln(0.01) \geq 0

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x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}

Where :

f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)

f'(x_n)=1-\frac{1}{1+\lambda}

Iterating as shown on the figure attached we find a final solution given by:

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