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3 years ago

Write an equation in point-slope form of the line that passes through the given point and has the given slope. (3, 2); m = 2/3

Mathematics

1 answer:

77julia77 [94]3 years ago

4 0

Answer:

$y - 2 = \frac{2}{3}(x-3)$

Step-by-step explanation:

Point-slope form: y - y1 = m(x - x1)

Slope: $\frac{2}{3}$

Point: (3, 2) = (x1, y1)

To write the equation in point-slope form, we need to know the values of the slope and one point. Since we've already been given those values, all we have to do is input them into the equation:

y - y1 = m(x - x1)

$y - 2 = \frac{2}{3}(x-3)$

The equation in point-slope form is: $y - 2 = \frac{2}{3}(x-3)$

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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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ivanzaharov [21]

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w = width of the rectangle
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Part A.
The area is
A = w*l
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Write the parabola in standard form.
A = -[l² - 2l]
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