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Irina-Kira [14]
2 years ago
3

your utility bill for april is $170. If you pay after the due date,a late payment panality of $7.72 is added? what is percent of

penality?
Mathematics
1 answer:
Galina-37 [17]2 years ago
5 0

Answer:

4.54%

Step-by-step explanation:

<u>Step 1: Write the data</u>

<em>Total bill = $170</em>

<em>Penalty = $7.72</em>

<em>Percentage of penalty = ?</em>

<u>Step 2: Write the formula to find the percentage of penalty</u>

<em>Percentage of penalty = Penalty/Total bill * 100</em>

<em>Percentage of penalty = 7.72/170 * 100</em>

Percentage of penalty = 4.54%

Therefore, the percent of penalty is 4.54%

!!

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Answer:

\left(y-8\right)

Step-by-step explanation:

y^2-10y+16

=\left(y^2-2y\right)+\left(-8y+16\right)

y^2-2y

=yy-2y

=y\left(y-2\right)

-8y+16

=-8y+8\cdot \:2

=-8\left(y-2\right)

=y\left(y-2\right)-8\left(y-2\right)

=\left(y-2\right)\left(y-8\right)

The other factor is:

\left(y-8\right)

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745436

Step-by-step explanation:

Maria has already written Two-fifths of her 1,000 word essay. If she continues writing at the same pace of 6One-half words per minute, which expression shows the amount of time it will take her to write the rest of the essay?

1000 times two-fifths times StartFraction 13 Over 2 EndFraction

1000 times two-fifths

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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:

(a) 0.0686

(b) 0.9984

(c) 0.0016

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,

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

\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)

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

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

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

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

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

The probability that the part is good (either defect free or slightly defective)

=\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

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

=\frac{0.9}{0.9014}

=0.9984

(c) The probability that the 'bad' (defective} parts get shipped

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

=1-0.9984

=0.0016

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