A falcon to fly 322 km/h how many meters per hour on the falcon fly

The answer to the problems and how to solve it.

swat32

Let x and y represent angles 1 and 2, respectively. The given relations tell you

... x + y = 180 . . . . the sum of measures of a linear pair is 180°

... y = 2x + 6 . . . . angle 2 is 6 more than twice angle 1

One way to solve these is to add twice the first equation to the second.

... 2(x + y) + (y) = 2(180) + (2x +6)

Now, subtract 2x

... 3y = 366

... y = 122 . . . . divide by the coefficient of y

The measure of angle 2 is 122°.

3 0

3 years ago

1. What is x squared over 6=w-y if the subject of the formula is x??? Help please

Kay [80]

$\frac{x^2}{6}=w-y\\
x^2=6(w-y)\\
x=\sqrt{6(w-y)} \vee x=-\sqrt{6(w-y)}$

8 0

2 years ago

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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

3 years ago

Plz Help!

Volgvan

The stack of cards are 1 inch tall all together taking up 8 inch’s (cubed)or half of the space

4 0

3 years ago

From the set {6, 7, 11}, use substitution to determine which value of x makes the equation true.

kotykmax [81]

11 if you multiply 11 by both the number and and the variable you’ll get 11x-55=66. Add 55 to each side which will give you 121 then divide that by 11

3 0

3 years ago

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