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OleMash [197]
3 years ago
6

Designer Dolls, Inc. found that the number N of dolls sold varies directly with their advertising budget A and inversely with th

e price P of each doll. The company sold 5200 dolls when $26,000 was spent on advertising and the price of a doll was set at $30. Determine the number of dolls sold when the amount spent on advertising is increased to $52,000. Round to the nearest whole number.
Mathematics
2 answers:
Hoochie [10]3 years ago
7 0

Answer:

The number of dolls sold 10,400

Step-by-step explanation:

Given:  The number N of dolls sold varies directly with their advertising budget A and inversely with the price P of each doll.

N = k(A/P), where k is the constant.

Now we have to find k.

Given: N = 5200, A = 26,000 and P = 30

5200 = k (26000/30)

5200 = k(866.67)

k = 5.99, when we round off we get k = 6

Now let's find the number of dolls sold when the ad amount increase to $52,000

Now plug k = 6, A = 52000 and p = 30

N = 6(52000/30)

N = (52000/5)

N = 10.400

Therefore, the number of dolls sold 10,400

Hope this will helpful.

Thank you.

GREYUIT [131]3 years ago
5 0
I think 104,000 is the answer because when 26,000 is double it's 52,000 so I doubled 52,000 and 104,000 is what I got.
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Step-by-step explanation:

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The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, ...

(a)

Compute the probability of the event (<em>X</em> > 84) as follows:

P (X > 84) = 1 - P (X ≤ 84)

                =1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007

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(b)

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P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)

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