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3 years ago

What is the greatest factor of 60 and 75

Mathematics

2 answers:

Radda [10]3 years ago

8 0

The greatest factor of 60 and 75 is 15.

Reasoning: I found the factors and prime factorization of 60 and 75. The biggest common factor number is the GCF number. So the greatest common factor 60 and 75 is 15.

Ugo [173]3 years ago

6 0

Answer:

Step-by-step explanation:

60 = 2 * 2 * 3 * 5 = 2^2 * 3 * 5

75 = 3 * 5 * 5 = 3 * 5^2

For the GCF, use common factors with lower exponent.

GCF = 3 * 5 = 15

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Your answer is $125$ miles

Step-by-step explanation:

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$10 \frac{5}{12}$ × $12$ = $125$

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3 years ago

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14 m

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3 years ago

In a widget factory, machines A, B, and C manufacture, respectively, 20, 30, and 50 percent of the total. Of their output 6, 3,

steposvetlana [31]

Answer:

38.71% probability it was manufactured by machine A.

29.03% probability it was manufactured by machine B.

32.26% probability it was manufactured by machine C.

Step-by-step explanation:

We have these following probabilities:

A 20% probability that the chip was fabricated by machine A.

A 30% probability that the chip was fabricated by machine B.

A 50% probability that the chip was fabricated by machine C.

A 6% probability that a chip fabricated by machine A was defective.

A 3% probability that a chip fabricated by machine B was defective.

A 2% probability that a chip fabricated by machine C was defective.

The question can be formulated as:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

What are the probabilities that it was manufactured by machines A, B, and C?

Machine A

What is the probability that the widget was manufactures by machine A, given that it is defective?

P(B) is the probability that it was manufactures by machine A. So $P(B) = 0.20$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine A. So $P(A/B) = 0.06$

P(A) is the probability that a widget is defective. This is the sum of 6% of 20%, 3% of 30% and 2% of 50%. So

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.2*0.06}{0.031} = 0.3871$

38.71% probability it was manufactured by machine A.

Machine B

What is the probability that the widget was manufactures by machine B, given that it is defective?

P(B) is the probability that it was manufactures by machine B. So $P(B) = 0.30$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine B. So $P(A/B) = 0.03$

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.03}{0.031} = 0.2903$

29.03% probability it was manufactured by machine B.

Machine C

What is the probability that the widget was manufactures by machine C, given that it is defective?

P(B) is the probability that it was manufactures by machine C. So $P(B) = 0.50$

P(A/B) is the probability that a widget is defective, given that it is manufactured by machine C. So $P(A/B) = 0.02$

$P(A) = 0.06*0.20 + 0.03*0.30 + 0.02*0.5 = 0.031$

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.02}{0.031} = 0.3226$

32.26% probability it was manufactured by machine C.

6 0

3 years ago