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3 years ago

Help please odds only geometry

Mathematics

2 answers:

yarga [219]3 years ago

5 0

$\log98=\log(7\cdot14)=\log 7+\log14=h+j\\\\
\log175=\log(5^2\cdot7)=\log5^2+\log7=2\log5+h=2i+h\\\\
\log2=\log\dfrac{14}{7}=\log14-\log7=j-h\\\\
\log245=\log(5\cdot7^2)=\log5+\log7^2=i+2\log7=i+2h$

nikklg [1K]3 years ago

3 0

1)
log 98 = log( 7 * 14 ) = log 7+ log 14= h + j ;

2)
log 175 = log ( 5^2 * 7 ) = log 5 ^ 2 + log 7 = 2 * log 5 + log 7 = 2i + h ;

3)
log 2 = log ( 14 / 7 ) = log 14 - log 7 = j - h ;

4)
log 245 = log ( 7^2 * 5 ) = log 7 ^2 + log 5 = 2 * log 7 + log 5 = 2h + i ;

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Answer:

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Step-by-step explanation:

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The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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x is the number of successes.

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$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Fleet of 18 means that $N = 18$

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9 are destroyed, which means that $n = 9$

What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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$P(X = 1) = h(1,18,9,9) = \frac{C_{9,1}*C_{9,8}}{C_{18,9}} = 0.001666$

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