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Damm [24]
3 years ago
13

(NO LINKS) I BEG U HALP FELLOW SMART PPL

Mathematics
1 answer:
bazaltina [42]3 years ago
3 0

Answer: See the image below for the filled out table.

The other root is x = -2

===========================================================

Explanation:

The turning point is at (1, -45) which is the vertex. This is where the graph goes downhill, and then turns around to go uphill, or vice versa. Depending on the direction, the vertex is the lowest point or the highest point on the parabola.

We have (h,k) = (1,-45) as the vertex, so h = 1 and k = -45

y = a(x-h)^2 + k

y = a(x-1)^2 + (-45)

y = a(x-1)^2 - 45

Now plug in any other point from the table. You cannot pick (1,-45) or else you won't be able to solve for the variable 'a'. Let's go for (0,-40)

We'll plug x = 0 and y = -40 into the equation above to solve for 'a'

y = a(x-1)^2 - 45

-40 = a(0-1)^2 - 45

-40 = a(-1)^2 - 45

-40 = a - 45

a-45 = -40

a = -40+45

a = 5

Therefore, the equation for this parabola is

y = 5(x-1)^2 - 45

As a way to check, we can plug in something like x = -3 to find that...

y = 5(x-1)^2 - 45

y = 5(-3-1)^2 - 45

y = 5(-4)^2 - 45

y = 5(16) - 45

y = 80 - 45

y = 35

Which matches what the table shows in the first column. I'll let you verify the other columns. As you can probably guess at this point, we'll plug in the x values to get the corresponding y values.

So for x = -2, we get...

y = 5(x-1)^2 - 45

y = 5(-2-1)^2 - 45

y = 5(-3)^2 - 45

y = 5(9) - 45

y = 45 - 45

y = 0

The result of 0 here indicates we have a root at x = -2. This is the other x intercept. The x intercept already given to us was x = 4.

The rest of the table is filled out using the same idea. You should get what you see below.

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Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

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\sigma = \sqrt{\frac{1-p}{p^{2}}

In this problem, we have that:

p = 0.383

So

\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61

\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05

(a) What is the probability that a drought lasts exactly 3 intervals?

This is f(3)

f(x) = (1-p)^{x}p

f(3) = (1-0.383)^{3}*(0.383)

f(3) = 0.09

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is P = f(0) + f(1) + f(2) + f(3)

f(x) = (1-p)^{x}p

f(0) = (1-0.383)^{0}*(0.383) = 0.383

f(1) = (1-0.383)^{1}*(0.383) = 0.236

f(2) = (1-0.383)^{2}*(0.383) = 0.146

Previously in this exercise, we found that f(3) = 0.09

So

P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4).

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

P(X \leq 3) + P(X \geq 4) = 1

0.855 + P(X \geq 4) = 1

P(X \geq 4) = 0.145

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