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alexgriva [62]
3 years ago
6

Or yk dont answer my question thats coo :/

Mathematics
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

whats the question i might be able to help

Step-by-step explanation:

:)))))))

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Help plssss ASAP brainliest
____ [38]

Answer:

Step-by-step explanation:

x and 6

6^2 is 36

x^2 is still x^2

now we need to find the square root of their sum

square root of ( 36 + x^2 )

I'm not that sure about this one i think it might be more simplified

8 0
2 years ago
PLZ HELP FAST
dolphi86 [110]
The answer is 2x-26
1+2x-27  9*3= 27
1-27+2x   You would combine like terms here
2x-26
8 0
3 years ago
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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
3 years ago
Solve this question
aleksklad [387]
The answer will be C 
6 0
3 years ago
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Jasper Diaz apostrophe Balance Sheet. Total assets are 15,800 dollars. Total liabilities are 4,400 dollars.
Thepotemich [5.8K]

Answer:

the total value of jaspers assets is:  $15,800

the total value of jaspers liabilities is: $4,400

Step-by-step explanation:

4 0
2 years ago
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