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4 years ago

Three consecutive even integers have a sum of 48 . find the integers.

Mathematics

1 answer:

Mandarinka [93]4 years ago

4 0

x+(x+2)+(x+4)=48

3x+6=48

Minus 6 to both sides

3x=42

Divide 3 to both sides

x=14

Get the next two integers

x,x+2,x+4 or 14,16,18 = 48 is you Answer

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The one you have selected is the right answer 126 times pi

Step-by-step explanation:

Exact answer - 395.840674

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3 years ago

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Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

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Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

0.95 probaiblity of hitting a target

This means that $p = 0.95$

10 targets

This means that $n = 10$

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

$P(X = 10) + P(X < 10) = 1$

We want P(X < 10). So

$P(X < 10) = 1 - P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987$

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3 years ago

Captain Ben has a ship, the H.M.S Crimson Lynx. The ship is five furlongs from the dread pirate Luis and his merciless band of t

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Step-by-step explanation:

In our case event A is pirate hitting captain's ship and event B is captain missing pirate's ship. We have been given that pirate shoots first so pirate's ship can't be hit before pirate shoots his cannons. So probability of hitting captain's ship is 1/3. We have been given that if Captain Ben's ship is already hit then Captain Ben will always miss. So the probability of Captain missing the dread pirate's ship given the pirate Luis hitting the Captain ship is 1. Now to find probability that pirate hits Captain, but Captain misses we will multiply our both probabilities.

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3 years ago

If g (x) = 1/x then [g (x+h) - g (x)] /h

lys-0071 [83]

Answer:

$\dfrac{-1}{x(x+h)}, h\ne 0$

Step-by-step explanation:

If $g(x) = \dfrac{1}{x}$ , then $g(x+h) = \dfrac{1}{x+h}$ . It follows that

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}$

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}$

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

$= \dfrac{-1}{x(x+h)}, h\ne 0$

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3 years ago

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