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3 years ago

6

20y^2-34y+6 (factoring)

1 answer:

denpristay [2]3 years ago

3 0

Hello There!

Your answer is 2(5y - 1)(2y - 3)

Hope This Helps You!
Good Luck :)

- Hannah ❤

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To solve for m in the formula , use the multiplication property of equality. 1)True or 2) False?

NeX [460]

True. just like Texaschic101 said DV= M

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3 years ago

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The diagram show a straight line ABC . Find the value of p

Ludmilka [50]

Finding the change in y/change in x=the slope, we get
(13-11)/(-4-(-2))=2/-2=-1. Therefore, the equation is y=-1*x+b. Plugging it into (-4, 13), we note that 13=4+b and b=9 by subtracting 4 from both sides, making the equation y=-x+9. Plugging it into A, we get that
1 (y , the second value) = -x+9. Subtracting both sides by 9, we get 1-9=-x and -8=-x. Multiplying each side by -1, we get x=8

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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

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

So

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

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 $P(A/B) = 0.03$

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

So

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

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 $P(A/B) = 0.02$

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

So

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

32.26% probability it was manufactured by machine C.

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3 years ago

Javier's monthly net income is $1,500 . He requests that his bank deposit 1/5 of his monthly earnings into his savings account .

icang [17]

Answer:

1500 * 1/5

1500 * 0.2 = 300

300 * 12 = 3600

Javier contributes $3,600 a year to his savings.

Step-by-step explanation:

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finlep [7]

$7 * \frac{3}{12} = \frac{7*3}{12}= \frac{21}{12}$

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3 years ago