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3 years ago

10

Solve this absolute value equation and show all work.

1 answer:

vitfil [10]3 years ago

6 0

I was confused at first until I realized that you'd shared not one, not two, but three questions in one post. would you please post just one question at a time to avoid this.

I'll focus on your second question only: Solve <span>3 + |2x - 4| = 15.

Subtr. 3 from both sides. Result: |2x - 4| = 12

Divide all terms by 2, to reduce: |x - 2| = 6

Case 1: x-2 is already +, so we don't need | |:

x - 2 = 6 => x = 8 (first answer)

Case 2: x-2 is negative, so |2x-4| = -(2x-4) = 6

Then -2x + 8 = 6. Subtr. 8 from both sides: -2x = -2

Div both sides by -2: x = 1 (second answer)

Be sure to check these results by subst. them into the original equation.

Please post your other questions separately. Thanks and good luck!

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Grasshoppers are distributed at random in a large field according to a Poisson process with parameter a 5 2 per square yard. How

HACTEHA [7]

In this question, the Poisson distribution is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Parameter of 5.2 per square yard:

This means that $\mu = 5.2r$ , in which r is the radius.

How large should the radius R of a circular sampling region be taken so that the probability of finding at least one in the region equals 0.99?

We want:

$P(X \geq 1) = 1 - P(X = 0) = 0.99$

Thus:

$P(X = 0) = 1 - 0.99 = 0.01$

We have that:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5.2r}*(5.2r)^{0}}{(0)!} = e^{-5.2r}$

Then

$e^{-5.2r} = 0.01$

$\ln{e^{-5.2r}} = \ln{0.01}$

$-5.2r = \ln{0.01}$

$r = -\frac{\ln{0.01}}{5.2}$

$r = 0.89$

Thus, the radius should be of at least 0.89.

Another example of a Poisson distribution is found at brainly.com/question/24098004

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