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3 years ago

I need help with this question

Mathematics

1 answer:

Setler [38]3 years ago

6 0

Answer:

Step-by-step explanation:

radius r = 4

height h = 8

volume = πr²h = π·4²·8 = 128π units³

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boyakko [2]

Answer:

warwar

Step-by-step explanation:

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3 years ago

Express cos^2 x in terms of cos2x. Hence, express f(x) = 6cos^2 x - 4sin 2x in the form R cos (2x + Y degrees) + p where R, y an

Lady bird [3.3K]

2cos^2 x = 1 + cos 2x
cos^2 x = 1/2(1 + cos 2x)
f(x) = 6cos^2 x - 4 sin 2x = 6(1/2(1 + cos 2x) - 4 sin 2x = 3 + 3cos 2x - 4sin 2x

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3 years ago

Which of the following is a unit for area?<br> ◦ m 3<br> ◦ inches<br> ◦ feet<br> ◦ cm 2

statuscvo [17]

Answer:

$\huge\boxed{Answer\hookleftarrow}$

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6 0

2 years ago

Read 2 more answers

Task: Nonpermissible Values for Rational Expressions [30 points] Write a rational expression that has the nonpermissible values

igor_vitrenko [27]

Answer:

A rational expression that has the nonpermissible values $x = 0$ and $x = 17$ is $f(x) = \frac{4}{x\cdot (x-17)}$ .

Step-by-step explanation:

A rational expression has a nonpermissible value when for a given value of $x$ , the denominator is equal to zero. In addition, we assume that both numerator and denominator are represented by polynomials, such that:

$f(x) = \frac{p(x)}{q(x)}$ (1)

Then, the factorized form of $q(x)$ must be:

$q(x) = x\cdot (x-17)$ (2)

If we know that $p(x) = 4$ , then the rational expression is:

$f(x) = \frac{4}{x\cdot (x-17)}$ (3)

A rational expression that has the nonpermissible values $x = 0$ and $x = 17$ is $f(x) = \frac{4}{x\cdot (x-17)}$ .

8 0

3 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:

$p = 0.383$

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

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

(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 $f(3) = 0.09$

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

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