Answer:
Step-by-step explanation:
My approach was to draw out the probabilities, since we have 3 children, and we are looking for 2 boys and 1 girl, the probabilities can be Boy-Boy-Girl, Boy-Girl-Boy, and Girl-Boy-Boy. So a 2/3 chance if you think about it, your answer 2/3 can't be correct. If we assume that boys and girls are born with equal probability, then the probability to have two girls (and one boy) should be the same as the probability to have two boys and one girl. So you would have two cases with probability 2/3, giving an impossible 4/3 probability for both cases. Also, your list "Boy-Boy-Girl, Boy-Girl-Boy, and Girl-Boy-Boy" seems strange. All of those are 2 boys and 1 girl, so based on that list, you should get a 100 percent chance. But what about Boy-Girl-Girl, or Girl-Girl-Girl? You get 2/3 if you assume that adjacencies in the (ordered) list are important, i.e., "2 boys and a girl" means that the girl was not born between the boys.
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<u>Sol</u><u>ution</u><u>:</u></h2>
Equation: x² + 10x + 21
<u>Step</u><u> </u><u>1</u><u>:</u> Find two numbers that can add up to 10 and be multiplied to 21. We have: 7 & 3, in the sense that 7+3=10, and 7×3=21. Replacing 10 with 7+3, the equation is now → x² + 7x + 3x + 21
<u>Step</u><u> </u><u>2</u><u>:</u> Get the new equation bracketed → (x² + 7x) (+3x + 21)
<u>Step</u><u> </u><u>3</u><u>:</u> Use 'x' in the equation. For the first part, we have 'x'. x² = x × x so, bring out one x out side the bracket, divide 7x by = 7 → x (x +7). Do the same for the second part by dividing 21 by 3 = 7, and then bringing out 3 from the bracket → 3 (x + 7).
Bringing everything together, we have: x(x+7) +3(x+7) → (x+3) (x+7)
<h3>
<u>Final</u><u> </u><u>ans</u><u>wer</u><u>:</u></h3>
(x+3) (x+7)
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Answer:
This is not a question. please provide a question.
