I ned help whit ligninger

Please answer the question below for 50 points. It is very easy. 6th-grade math!

Korvikt [17]

Answer:

B

I might be wrong, please don't yell at me if I am :(

5 0

2 years ago

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Need help with this geometry question

Genrish500 [490]

Answer:

I think it's 6 (square root) 2

5 0

2 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.

6 0

2 years ago

SOMEONE HELPPP THANKYOUU

Eduardwww [97]

Answer:

5(8 - 2t) = 2 + 16(4 + t)

open the brackets;

use 5 to mulitiply (8 - 2t) = 40 - 10t

use 16 to mulitiply (4 + t) = 64 + 16t

then we have 40 - 10t = 2 + 64 + 16t

collect like terms

40 - 2 - 64 = 16t + 10t

-26 = 26t

divide both sides by 26

t = -1

5 0

2 years ago

PLEASE HELP GIVING BRAINEST AWNSER AND ITS WORTH 85 points

Dennis_Churaev [7]

Answer:

H. 10

Step-by-step explanation:

We can write down all the games that can be played (I'm going to use a, b, c, d, and e to symbolize the players)

ab, ac, ad, ae, bc, bd, be, cd, ce, de

As you can see, there are 10 games that are going to be played.

Best of Luck!

5 0

2 years ago

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