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3 years ago

Simplify this equation

Mathematics

1 answer:

Sati [7]3 years ago

5 0

Answer:

60√3

Step-by-step explanation:

All we are doing here is multiplying:

12√75

60√3

Alternatively, we could plug this into a calc and figure out our answer.

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Simplify 45 - 8 + 12 - (-25)

Mrac [35]

Answer:

Step-by-step explanation:

BODMAS

12-(-25)

=12+25=37

45-8+37

45-8=37

37+37=74

=74

4 0

3 years ago

Read 2 more answers

– 5 | 3 z + 8| –5 = -20

olchik [2.2K]

I'm assuming you want to solve for z :)

By simplifying both sides of the equation, then isolating the variable.

Exact form:

z= -5/3 , - 11/3

Mixed number form:

z= −1 2/3 , −3 2/3

8 0

3 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:

$p = 0.383$

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

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

(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

Please Help me answer this!!!!!!

xxMikexx [17]

Answer:

yes

Step-by-step explanation:

4 0

3 years ago

Will mark brainliest :) Given vectors u = -9i + 8j and v = 7i + 5j find 2u -6j in terms of unit vectors i and j

a_sh-v [17]

Answer:

Step-by-step explanation:

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3 years ago

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Simplify this equation​

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