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bija089 [108]
3 years ago
14

I need help with this question pls anybody?​

Mathematics
1 answer:
HACTEHA [7]3 years ago
4 0

Answer:

The second one, the third one.

Step-by-step explanation:

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y value is-8

x value is-8✓2

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Solve:<br> 2(2x + 3) = -6(x + 9)
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Answer:

The answer is x = -6

Step-by-step explanation:

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Read 2 more answers
There are 76 men and 69 women in a room. 29 of the men and 23 of the women earn more than $50,000 per year.
AlladinOne [14]

Answer:

  52/145 ≈ 0.359 = 35.9%

Step-by-step explanation:

There are a total of 76+69 = 145 people in the room. Of them, 29+23 = 52 earn more than $50,000 per year. The probability of choosing one of those at random is 52/145 ≈ 0.35862 ≈ 35.9%.

___

For all of these probability problems, the probability of selecting an item with particular characteristics is the ratio of the number of such items to the total number of possible choices.

3 0
3 years ago
Frank chooses a box at random. If boxes 1 and 3 are chosen, Frank extracts 2 balls without replacement. If box 2 is chosen Frank
mars1129 [50]

Complete question:

Consider 3 boxes, each of which contains 4 balls in particular, box 1 contains 4 white balls, box 2 contains 3 white balls and 1 red ball, box 3 contains 2 white balls and 2 red balls. Frank chooses a box at random. If boxes 1 and 3 are chosen, Frank extracts 2 balls without replacement. If box 2 is chosen Frank extracts 2 balls with replacement. USE ONLY THE CONDITIONAL PROBABILITY FORMALISM and derive an expression for the probability that the 2 extracted balls are red.

Answer:

11/288

Step-by-step explanation:

We are given:

Box 1: ( 4White, ORed)

Box 2: (3White, 1Red)

Box 3: (2White, 2 Red)

We are told that if Frank choose from Box 1 and Box 3, 2 balls are extracted without replacement.

Since there is no red ball in Box 1, there is no way 2 red balls will come out from Box1.

Our Event, E = getting 2 red balls.

Now Box 1 is ruled out, we have:

P[E(B1)]= 0

P[E/B3)] = (2 2) / (4 2)

= 1/6

If box 2 is chosen, 2 balls are extracted with replacement. Therefore for Box 2:

P(E/B2) = (1/4) *(1/4)

= 1/16

Therefore probability that 2 balls extracted are red, we have:

P(E)=P(E/B1) P(B1) + P(E/B2) P(B2)+P(E/B3) P(B3)

= 0 * \frac{1}{3} + \frac{1}{6}*\frac{1}{3} + \frac{1}{16}*\frac{1}{3}

= 11/288

3 0
3 years ago
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