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3 years ago

URGENT URGENT URGENT!!! HELP PLEASE!!

Mathematics

2 answers:

Fynjy0 [20]3 years ago

8 0

Answer:

I think it is b im pretty sure

viva [34]3 years ago

3 0

Answer:

Step-by-step explanation:

I think it’s b correct me if I’m wrong

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Lola measures 3 buttons for a shirt cheese making one button is 1/8 inch 1 is 3/8 inch and 1 is 1/4 inch which button is smalles

Solnce55 [7]

Step-by-step explanation:

no se incluye la información que le envié en el mundo de los incas y la información de las personas en el mismo tiempo para la actividad que no me han llegado y que me digan que me digan que es

7 0

3 years ago

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

3 years ago

Help asap, please!!!!!!!!

Artemon [7]

1/6p + (-4/5) is the equivalent expression. You have to add like terms, meaning constants are added to constants, variables are added to variables, etc. the result you get from adding like variables leaves you with 1/6p + (-4/5) or 1/6p - 4/5

8 0

3 years ago

If f(x) = 2x2 + 2, find f(5).

nalin [4]

answer for this question 22 if f(5)

6 0

3 years ago

What is the value of n In the following 100 : 500 = n : 5

mrs_skeptik [129]

Step-by-step explanation:

ratio means over so,

100 = n

500 5

500n = 500

500

n=1

6 0

3 years ago