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2 years ago

1. The number of employees at a certain company is 1440 and is increasing

Mathematics

1 answer:

adelina 88 [10]2 years ago

6 0

Answer:

$y = 1440 e^{0.015 t}$

$y(9) = 1440 e^{0.015*9}= 1648.133$

So then after 9 years we will have approximately 1649 number of employees

Step-by-step explanation:

For this case we want to model the number of employees and we need to use an exponential model given by this general expression:

$y = y_o e^{rt}$

For this case the initial amount is $y_o = 1440$ and the rate $r =0.015$

And then the model would be given by

$y = 1440 e^{0.015 t}$

And if we find the value for t =9 years we got:

$y(9) = 1440 e^{0.015*9}= 1648.133$

So then after 9 years we will have approximately 1649 number of employees

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A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

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Answer:

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(b) 0.9984

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

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$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

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Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

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$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

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$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

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=1- the probability that the 'good' parts get shipped

=1-0.9984

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Answer:

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Step-by-step explanation:

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we all know them so we replace:

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b) Now, to know the annual savings, we must calculate the difference between the expense of the normal flashlight and the LED, let's calculate the daily cost of the LED, like this:

20 watt * 1 kw / 1000 watt = 0.02 kw

Daily cost = 0.02 kw * 12 h * 11 cents / kwh = 2.64 cents

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