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3 years ago

10

3/x+2 = 2/x−1 pleasee help

1 answer:

olga_2 [115]3 years ago

7 0

So you have:
$\frac{3}{x} + 2 = \frac{2}{x} - 1
$

1) Start off by multiplying both sides by x to simplify the problem:
$\frac{3}{x} + 2 = \frac{2}{x} - 1 \\
x(\frac{3}{x} + 2) = (\frac{2}{x} - 1)x\\
3 + 2x = 2 - x
$

2) Now isolate the x by moving all terms with x onto one side and all other terms onto the other side:
$3 + 2x = 2 - x\\
3x = -1$

3) Finally, divide by 3 on both sides to find the value of x:
$3x = -1\\
x = \frac{-1}{3}$

------

Answer: x = $\frac{-1}{3}$

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3 years ago

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

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Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

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

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

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ 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 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

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

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

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

So

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Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

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