2(x - y) in its distributed form is

Find the center and radius of the circle with equation (x + 7) + (y + 5)2 = 81.

vesna_86 [32]

Answer:

rtjgdesss eeevywxx13 46

7 0

4 years ago

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

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:

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

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

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

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:

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

$0.855 + P(X \geq 4) = 1$

$P(X \geq 4) = 0.145$

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

8 0

4 years ago

MAY SOMEONE PLZ HELP ME ON THIS, I DONT KNOW THE EQACTION

just olya [345]

Answer:

I will use "complete-the-square" approach if thats ok

first 0.5(x - 2)² - 2

second -2(x - 4)² + 8

7 0

2 years ago

How many hours pass between the two temperatures shown?

Blizzard [7]

Answer:

7 . time os on axis X. total vector lenght is 7.

8 0

2 years ago

Heeeeeeeeeeeeeeeeeeepl

DIA [1.3K]

Answer:

Hi there!

Your answer is;

a)

i) 400% of 240

240 is 100%

× 4

960= 400%

ii) 40% of 240

100% = 240

/100

1% = 2.4

× 40

40% = 96

iii) 4% of 240

100% is 240

/100

1% = 2.4

× 4

4% = 9.6

iv) .04% of 240

100% = 240

/100

1% = 2.4

/100

.01% = .024

× 4

.04% = .096

b) the patterns is that all these numbers equal sometime 96. Each of these have a different decimal place, but have the same actual numbers.

c) 4000% = 240

take the pattern:

400% is 960

Scale it up to 4000 by 10

400% is 960

× 10

4000% is 9600

Hope this helps!

5 0

3 years ago