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Gala2k [10]
3 years ago
5

I need help ASAP PLEASEE

Mathematics
1 answer:
Alex787 [66]3 years ago
4 0

Answer:

9. 21

10. 70

11. 64

12. 31

13. 85

14. 31

double check the answer, i didnt use a calculator so might be wrong

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I believe your answer is b.
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How to put this in $799.73 in numeric please help thank you
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The answer is 800$...
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Divide f(x) by d(x), and write a summary statement in the form indicated.
lukranit [14]
( x^4 - 4 x³ + 2 x² - 4 x + 1 ) : ( x² + 1 )   =  x² - 4 x + 1
- x^4             - x²
------------------------
          - 4 x³ + x² - 4 x
            4 x³        + 4 x
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( x^4 - 4 x³ + 2 x² - 4 x + 1) =  ( x² + 1 ) ( x² - 4 x + 1 )
3 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
BRAINLIEST ANSWER GIVEN IF YOU ARE CORRECT
GuDViN [60]

Answer:

a b c e

Step-by-step explanation:

4 0
3 years ago
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