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kolbaska11 [484]
3 years ago
11

Please help solve for b.

Mathematics
2 answers:
Zanzabum3 years ago
6 0
Subtract a from both sides of the equation.

b=−a+14

katen-ka-za [31]3 years ago
3 0

b + a = 14

b = 14 - a


14 - a + a = 14

14 = 14


Since we can't gain any further information after solving for b the first time, we know b = 14 - a to be the simplest form of b.

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Step-by-step explanation:

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Let h(x) = x*2 + kw - 8. If h(-1) = -5, what is the value of h(4)?<br> Help
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A jar contains two blue and five green marbles. A marble is drawn at random and then replaced. A second marble drawn at random.
DerKrebs [107]

corrected question:A jar contains two blue and five green marbles. A marble is drawn at random and then replaced. A second marble drawn at random. For each of the following, find the probability that: a) both marbles are blue b) both marbles are the same color  c)the marbles are different in color

Answer:

Step-by-step explanation:

<u><em>The probability was done with replacement.</em></u>

Probaility of an event happening=\frac{number of required outcomes}{number of possible outcomes}

number of blue marbles= 2

number of green marbles=5

total number of marbles=7

(a) probability that both marbles are blue = pr(first is blue)*pr(second is blue)

                                  =\frac{2}{7}*\frac{2}{7}

                                   =\frac{4}{49}

(b)probability that both marbles are the same color =pr(first is blue)*pr(second is blue) + Pr(first is green)*Pr(second is green)

   =\frac{4}{49} + \frac{5}{7}*\frac{5}{7}

   =\frac{4}{49} + \frac{25}{49}

 =\frac{29}{49}

(c)Probability that the marbles are different in colors=pr(first is blue)*pr(second is green) + Pr(first is green)*Pr(second is blue)

  =\frac{2}{7}*\frac{5}{7}+ \frac{5}{7}*\frac{2}{7}

           =\frac{10}{49}+\frac{10}{49}

      =\frac{20}{49}

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3 years ago
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