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4 years ago

Donte simplified the expression below.

2 answers:

8 0

Based on the given solution above made by Donte, the mistake that he made is that he did not apply the distributive property correctly for 4(1 + 3i). The answer is option A. Here is why. If we simplify this, it would look like this:
4(1 + 3i) – (8 – 5i)
4+12i - 8 + 5i
17i-4
so, 4(1 + 3i) – (8 – 5i) should be equal to 17i - 4.
Hope this is the answer that you are looking for. Have a great day!

sergij07 [2.7K]4 years ago

6 0

The answer is: [A]: He did not apply the distributive property correctly for 4(1 + 3i) .
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Explanation:
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Note the distributive property of multiplication:
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a*(b+c) = ab + ac.
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As such: 4*(1 + 3i) = (4*1) + (4*3i) = 4 + 12i ;
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Instead, Donte somehow incorrectly calculated:
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4*(1 + 3i) = (4*1) + 3i = 4 + 31; (and did the rest of the problem correctly);

Note: - (8 - 5i) = -8 + 5i (done correctly;
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So if Donte did not apply the distributive property correctly for 4*(1+3i)—and incorrect got 4 + 3i (as mentioned above); but did the rest of the problem correctly, he would have got:
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4+ 3i - 8 + 5i = -4 + 8i (the incorrect answer as stated in our original problem.
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This corresponds to: "Answer choice: [A]: He did not apply the distributive property correctly for 4(1 + 3i)."
___________________________

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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

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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 $\lambda$

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

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:

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