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3 years ago

11

one saturday at the farmers market each of the 94 vendors made 502 dollars in profit how much profit did all vendors make that s

aturday

2 answers:

Marina CMI [18]3 years ago

6 0

Rhe answer to that is 47188

soldier1979 [14.2K]3 years ago

3 0

The answer is 502×94=47188

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1 7/100

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Select the correct answer. If the function 5x + y = 1 has the domain (-2, 1, 6), then what is the corresponding range? OA. {-9,

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D) {11,-4,-29}

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3 years ago

Andrew must spend less than 53$ on meals during the weekend. he has already spent 21$ on meals costing 8$ average. how many addi

Shkiper50 [21]

Andrew can buy maximum 3 meals this weekend.

From given question,

Andrew must spend less than 53$ on meals during the weekend.

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8x + 21 < 53

We need to find the number of meals he can buy this weekend.

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1 year ago

HELP , TIMES ALMOST UP!!!!!!

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2 years ago

The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine

salantis [7]

The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

$f(x)=\frac{e^{\frac{-x}{1000} }}{1000}$

for x > 0. Determine the probability that

a. A component lasts more than 3000 hours before failure.

b. A componenet fails in the interval from 1000 to 2000 hours.

c. A component fails before 1000 hours.

d. Determine the number of hours at which 10% of all components have failed.

Answer: a. P(x>3000) = 0.5

b. P(1000<x<2000) = 0.2325

c. P(x<1000) = 0.6321

d. 105.4 hours

Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = $\int\limits^b_a {P(x)} \, dx$

Then, for the electronic component, probability will be:

P(a<x<b) = $\int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx$

P(a<x<b) = $\frac{1000}{1000}.e^{\frac{-x}{1000} }$

P(a<x<b) = $e^{\frac{-b}{1000} }-e^\frac{-a}{1000}$

a. For a component to last more than 3000 hours:

P(3000<x<∞) = $e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}$

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = $e^{-3}$

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = $e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}$

P(1000<x<2000) = $e^{-2}-e^{-1}$

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = $e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}$

P(0<x<1000) = $e^{-1}-1$

P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = $e^{\frac{-b}{1000} }-e^\frac{-a}{1000}$

0.1 = $1-e^\frac{-x}{1000}$

$-e^{\frac{-x}{1000} }=-0.9$

${\frac{-x}{1000} }=ln0.9$

-x = -1000.ln(0.9)

x = 105.4

10% of the components will have failed at 105.4 hours.

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3 years ago