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Serggg [28]
3 years ago
13

What would x equal for these two? 15/24 = x/3 & 7/24 = 3/x Thank you!!!

Mathematics
1 answer:
vesna_86 [32]3 years ago
6 0

The first one you have \frac{15}{24}=\frac{x}{3} . To solve this you would multiply both sides by 3.

\frac{45}{24}=x

Then simplify by dividing the numerator and denominator by 3.

\frac{15}{8}=x (or in a mixed fraction 1\frac{7}{8})

The second one you have \frac{7}{24}=\frac{3}{x} . To solve this you would multiply both sides by x.

\frac{7}{24}x=3

Now you would divide both sides by \frac{7}{24}.

x=\frac{3}{\frac{7}{24}}

You now then multiply 3 by 24 and then divide by 7.

\frac{72}{7}

x=\frac{72}{7} (in a mixed fraction would be 10\frac{2}{7})

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Suppose that 7 in every 10 auto accidents involve a single vehicle. If 15 auto accidents are randomly selected, compute the prob
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\mathbf{P(X \le 4 ) \simeq 0.0006722}

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P(X\le 4) \\ \\P(X \le 4) = P(X = 0) + P(X =1 ) + ... + P(X = 4)

P(X \le 4 ) = (^{15}_0) *0.7^0 *0.3^{15-0} + (^{15}_1) *0.7^1 *0.3^{15-1} + (^{15}_2) *0.7^2 *0.3^{15-2} + (^{15}_3) *0.7^3 *0.3^{15-3} + (^{15}_4) *0.7^4 *0.3^{15-4}

P(X \le 4 ) = (\dfrac{15!}{0!(15-0)!}) *0.7^0 *0.3^{15-0} + (\dfrac{15!}{1!(15-1)!}) *0.7^1 *0.3^{15-1} + (\dfrac{15!}{2!(15-2)!}) *0.7^2 *0.3^{15-2} + (\dfrac{15!}{3!(15-3)!}) *0.7^3 *0.3^{15-3} + (\dfrac{15!}{4!(15-4)!}) *0.7^4 *0.3^{15-4}

P(X \le 4 ) =1.4348907 \times 10^{-8} +5.02211745 \times 10^{-7} + 8.20279183 \times 10^{-6} + 8.29393397 \times 10^{-5} + 5.80575378  \times 10^{-4}

P(X \le 4 ) =6.7223407 \times 10^{-4}

\mathbf{P(X \le 4 ) \simeq 0.0006722}

3 0
3 years ago
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