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coldgirl [10]
3 years ago
7

Mark shoots arrows at a target and hits the bull's-eye about 40% of the time. find the probability that he hits the bull's-eye o

n the fourth shot
Mathematics
1 answer:
tiny-mole [99]3 years ago
4 0
The answer to this question would be: 40% or 0.4

In this question, you are asked the probability of hitting bull's eye on the fourth shot. This event is only one(only fourth, not all four shot) and should be independent to the first 3 shots. That mean, the chance should still 40% since every shot chance is 40%.
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\bold{r+\frac{3Q}h=t}\\-r\qquad\ -r\\\bold{\frac{3Q}h=t-r}\\\cdot h\qquad \cdot h\\ \bold{3Q=(t-r)h}\\^{\div(t-r)\quad\div(t-r)}\\\bold{\dfrac{3Q}{t-r}=h}

Or, if you mean (r+3Q)/h=t:

\bold{\frac{r+3Q}h=t}\\{}\ \ \cdot h\quad\ \cdot h\\\bold{r+3Q=ht}\\{}\ \ \div t\qquad \div t\\\bold{\dfrac{r+3Q}{t}=h}\\\\\bold{h=\dfrac{r+3Q}{t}}

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