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olasank [31]
2 years ago
10

Which of the following options have the same value as %5, percent of 35? Choose 2 answers: (Choice A) A 5*35 (Choice B) B 5/100*

35 (Choice C) C 0.5* 0.35 (Choice D) D 0.05*35 (Choice E) E 5/10*35
Mathematics
2 answers:
nasty-shy [4]2 years ago
3 0

Answer:

B and D

Step-by-step explanation:

5% of 35 can be represented as 5% \cdot35.

To find equations that make this work, we need to find different ways to represent 5%.

5% can be represented as fraction - over 100. The percentage over 100 will be equal to the percent.

So, \frac{5}{100}. Multiplying by 35 gets us \frac{5}{100}\cdot35

This means that Choice B is correct.

Another way to represent 5% is as a decimal.

We already know that the fraction form of 5% is \frac{5}{100}. This means that in the decimal, the number 5 will be two place values to the right of the decimal place.

0.<u>0</u><u>5</u>

So the decimal expansion of 5% is 0.05. Multiplyin by 35 get us 0.05\cdot35, so choice D works too.

Hope this helped!

aniked [119]2 years ago
3 0

Answer:

<h3>The answer is option D and B</h3>

Step-by-step explanation:

To find the the value of the percentage of a number divide the percentage by 100 and multiply it by the number

So we have

5 percent of 35 is written as

\frac{5}{100}  \times 35 \\  =  \frac{1}{20}  \times 35 \\  =  \frac{7}{4}

We have the answer as

1.75

From the options the one that has the same value as the above answer are

0.05 × 35 and 5/100 × 35

Hope this helps you

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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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(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

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}}

=\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

5 0
2 years ago
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