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4 years ago

Solve this inequality:−9x−6≥−3(x−4)

2 answers:

8 0

X ≤ − 3; First solve for the (), so −9x−6≥−3x+12. Then you combine Add 3x to both sides: −9x−6+3x≥−3x+12+3x−6x−6≥12
Same for 6 : −6x−6+6≥12+6−6x≥18
Divide both sides by -6; Your done!

Ratling [72]4 years ago

4 0

X ≥ 0 is the answer for -9x-6≥ -3(x-4)

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Reil [10]

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3 years ago

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Ms. Stevens gave a 30 question quiz to her Algebra 1 class. After grading the quizzes, her students received the following score

Lunna [17]

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4 years ago

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Lera25 [3.4K]

Answer:

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3 years ago

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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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3 years ago

Can someone please answer this!

Aleksandr [31]

Answer:
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4 years ago