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azamat
3 years ago
5

Please answer this question!!thanks xx

Mathematics
1 answer:
Simora [160]3 years ago
7 0
180-(180-49-48)
180-(83)
97

180-((180-141)+(180-76))
180-(39+104)
180-143
137

180-((180-101)+60)
180-(79+60)
180-139
41

360-147-79
213-79
139

180-(360-109*2)
180-(360-218)
180-142
38

(180-118)/2
62/2
31

Hope this helps :)
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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
ludmilkaskok [199]

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

0.01 \geq e^{-\lambda}(1+\lambda)

Applying natural log on both sides we have:

ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)

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

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

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

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:

\lambda \geq 6.63835

4 0
2 years ago
Simplify the expression (2^-4)^-2
Vedmedyk [2.9K]

Answer:

256

Step-by-step explanation:

Hope this helps!

If not, I am sorry.

6 0
1 year ago
Which is the domain of the function f(x) = -5/6(3/5)^x?
lutik1710 [3]

the domain of f is {\bf R} = (-\infty, \infty)

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3 years ago
How to simplify 3sin^2x + 3(1-sin^2x)
EleoNora [17]

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Lubov Fominskaja [6]

Answer:

y-4=-7(x-3) and y=-3/8x-8

Step-by-step explanation:

Point slope form: y-y1=m(x-x1)

the slope is already -7.

We can the just plug in the values: y-4=-7(x-3)

For the 2nd problem we can plug in values as well: -2=3/8(16)+b.

-2=6+b. the y intercept is -8. The final form is y=-3/8x-8

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