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rjkz [21]
3 years ago
7

Help me!!!!!! I’ll been stuck on this for to long

Mathematics
2 answers:
hodyreva [135]3 years ago
7 0

The distribution is very simple. Using FOIL.

It states that (a+b)(c+d)=ac+ad+bc+bd.

Also note that when multiplying expressions we multiply variables and values differently. If we have variables like x their exponents will add. If we have values like 3 we multiply them normally.

For example your practise 3.

(x+3)(2x^2+4)=2x^2\cdot x+4x+3\cdot2x^2+3\cdot4 \\\underline{2x^3+4x+6x^2+12}

Now just order the expressions from bigger exponent to smaller and than values. (Usual notation although no need).

And solution is:

\boxed{2x^3+6x^2+4x+12}

Hope this helps.

r3t40

posledela3 years ago
6 0

Answer:

See below

Step-by-step explanation:

a\cdot(b + c) = a\cdot b + a\cdot c

3) Practice: Organizing information

\begin{array}{lll}\qquad \textbf{Steps} & \textbf{Problem: }(x + 3)(2x^{2} + 4) & \\\textbf{1. List variables} & a = x + 3 & \\ & b = 2x^{2} & \\ & c = 4 &\\\\\textbf{2. Distribute} & (x + 3)(2x^{2} + 4)& = (x + 3)(2x^{2}) + (x + 3)(4)\\\\\textbf{3. Redistribute} & (x + 3)(2x^{2})& (x + 3)(4)\\& a = 2x^{2} & a = 4\\& b = x & b = x\\& c = 3 & c = 3\\& 2x^{3} + 6x^{2} & 4x + 12\\\textbf{4. Combine}& & \\\qquad\textbf{terms} & 2x^{3} + 6x^{2}+ 4x + 12 & \\\end{array}

4. Practice: Summarizing

\text{You can use the FOIL method to multiply two }\underline{\text{binomials}}.\\\text{The letters in FOIL stand for }\underline{\text{First, Outer, Inner, Last}}.\\\text{The FOIL method helps you to remember how to multiply each term in one }\\\underline{\text{binomial}} \text{ by each term in the other }\underline{\text{binomial}}.

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Answer:

B = 1.875

Step-by-step explanation:

given that A varies directly as B and inversely as C then the equation relating them is

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when A = 10 and C = 1.5 , then

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And the probability of loss with the first wersion is 0.729

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Solution to the problem

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Let X the random variable of interest, on this case we now that:

X \sim Binom(n=3, p=0.1)

The probability mass function for the Binomial distribution is given as:

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Where (nCx) means combinatory and it's given by this formula:

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We can find the probability of loss like this P(X=0) and if we find this probability we got this:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

Alternative 2

Let Y the random variable of interest, on this case we now that:

Y \sim Binom(n=5, p=0.05)

The probability mass function for the Binomial distribution is given as:

P(Y)=(nCy)(p)^y (1-p)^{n-y}

Where (nCx) means combinatory and it's given by this formula:

nCy=\frac{n!}{(n-y)! y!}

We can find the probability of loss like this P(Y=0) and if we find this probability we got this:

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

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Step-by-step explanation:

smartness bc i love math

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