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Hatshy [7]
3 years ago
8

The owner of a small deli is trying to decide whether to discontinue selling magazines. He suspects that only 9.8% of his custom

ers buy a magazine and he thinks that he might be able to use the display space to sell something more profitable. Before making a final decision, he decides that for one day he will keep track of the number of customers that buy a magazine.(b) Assuming his suspicion that 9.8% of his customers buy a magazine is correct, what is the probability that exactly 2 out of the first 11 customers buy a magazine? Give your answer as a decimal number rounded to two digits.(c) What is the expected number of customers from this sample that will buy a magazine?
Mathematics
1 answer:
NISA [10]3 years ago
3 0

Answer:

Step-by-step explanation:

Let x be a random variable representing the number of his customers that buy a magazine. This is a binomial distribution since the outcomes are two ways. It is either a selected customer buys a magazine or he doesn't. The probability of success, p = 9.8/100 = 0.098

The probability of failure, q would be 1 - p = 1 - 0.098 = 0.902

b We want to determine P(x = 2)

n = 11

From the binomial distribution calculator,

P(x = 2) = 0.21

c) the expected number of customers from this sample that will buy a magazine is same as the mean.

mean = np

mean = 11 × 0.098 = 1.08

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8 0
3 years ago
What is the product in simplest form x^2+9x+18/x+2 times x^2-3x-10/x^2+2x-24
german
\dfrac{x^2+9x+18}{x+2}\cdot\dfrac{x^2-3x-10}{x^2+2x-24}=\dfrac{x^2+6x+3x+18}{x+2}\cdot\dfrac{x^2-5x+2x-10}{x^2+6x-4x-24}\\\\=\dfrac{x(x+6)+3(x+6)}{x+2}\cdot\dfrac{x(x-5)+2(x-5)}{x(x+6)-4(x+6)}

=\dfrac{(x+6)(x+3)}{x+2}\cdot\dfrac{(x-5)(x+2)}{(x+6)(x-4)}=\dfrac{(x+3)(x-5)}{x-4}\\\\=\dfrac{x^2-5x+3x-15}{x-4}=\dfrac{x^2-2x-15}{x-4}
5 0
3 years ago
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Which of the following is the solution to the differentiable equation dy/dx=4x/y, where Y(2)=-2
aliya0001 [1]

First we'll substitute \frac{dy}{dx} with y'

y'=\frac{4x}{y}

Then we can separate this.

yy'=4x

Then we'll solve this.

yy'=4x: \frac{y^2}{2}=2x^2+c_1

\frac{y^2}{2}=2x^2-6

y=2\sqrt{x^2-3},y=-2\sqrt{x^2-3}

Then we'll plug in to find the extraneous solutions (if any)

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7 0
3 years ago
Memory module consists of 9 chips. The device is designed with redundancy so that it works even if one of its chips is defective
soldier1979 [14.2K]

Answer:

a) P[C]=p^n

b) P[M]=p^{8n}(9-8p^n)

c) n=62

d) n=138

Step-by-step explanation:

Note: "Each chip contains n transistors"

a) A chip needs all n transistor working to function correctly. If p is the probability that a transistor is working ok, then:

P[C]=p^n

b) The memory module works with when even one of the chips is defective. It means it works either if 8 chips or 9 chips are ok. The probability of the chips failing is independent of each other.

We can calculate this as a binomial distribution problem, with n=9 and k≥8:

P[M]=P[C_9]+P[C_8]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)\\\\P[M]=p^{8n}(p^{n}+9(1-p^n))\\\\P[M]=p^{8n}(9-8p^n)

c)

P[M]=(0.999)^{8n}(9-8(0.999)^n)=0.9

This equation was solved graphically and the result is that the maximum number of chips to have a reliability of the memory module equal or bigger than 0.9 is 62 transistors per chip. See picture attached.

d) If the memoty module tolerates 2 defective chips:

P[M]=P[C_9]+P[C_8]+P[C_7]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1+\binom{9}{7}P[C]^7(1-P[C])^2\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])+36P[C]^7(1-P[C])^2\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)+36p^{7n}(1-p^n)^2

We again calculate numerically and graphically and determine that the maximum number of transistor per chip in this conditions is n=138. See graph attached.

6 0
4 years ago
One cases of books weight 35 pounds.One case of magazines weights 23 pounds.A book store wants to ship 72 cases of books and 94
schepotkina [342]
I'm sorry but it's bugging me that you aid 'one cases', you can't have the plural of something that is strictly singular.

Now: for your answer.
72*35=2520
94*23=2162
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The total wight is 4682 pounds.

Hope this helps
6 0
3 years ago
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