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2 years ago

8

shania gets $40 per day as wages and $4.50 as commission for every pair of shoes he sells in a day. His daily earnings goal is $

112.

1 answer:

forsale [732]2 years ago

3 0

Yeah, dude it sure is.

I'm assuming you wanted something like this?

40x + 4.5y = 112

x = Days

y = Pairs of shoes

112 = Total profits

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3 years ago

Mean for 16, 12, 13, 22

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3 years ago

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If you notice the picture below, the amount of fencing, or perimeter, that will be used will be 3w + 2l

now $\bf \begin{cases}
A=l\cdot w\\\\\
A=3000000\\
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\frac{3000000}{w}=l
\end{cases}\qquad thus
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P=3w+2l\implies P=3w+2\left( \cfrac{3000000}{w} \right)\implies P=3w+\cfrac{6000000}{w}
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P=3w+6000000w^{-1}\\\\
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now\qquad \cfrac{dP}{dw}=3-\cfrac{6000000}{w^2}\implies \cfrac{dP}{dw}=\cfrac{3w^2-6000000}{w^2}
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3 years ago

Let X1 and X2 be two random variables following Binomial distribution Bin(n1,p) and Bin(n2,p), respectively. Assume that X1 and

ryzh [129]

Answer:

a) X1+X2 have distribution Bi(n1+n2, p)

b)

P(X1+X2 = 1 | X2 = 0) = np(1-p)ⁿ¹⁻¹

P(X1+X2 = 1| X2 = 1) = (1-p)ⁿ¹

P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

Step-by-step explanation:

Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).

b)

Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).

If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹

If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹

Thus,

P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

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3 years ago