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valentinak56 [21]
3 years ago
9

In mathematics given that x=L2-L1/L1(O2-O1).

Mathematics
1 answer:
Assoli18 [71]3 years ago
4 0

Answer:

1. L1 = \frac{L2}{X(O2 -O1)}

2. O2 = \frac{L2 - L1(1 + XO1)}{XL1}

Step-by-step explanation:

Given: x = \frac{L2 - L1}{L1(O2 -O1)}

1. To make L1 the subject of formula;

x = \frac{L2 - L1}{L1(O2 -O1)}

cross multiply to have;

L2 - L1 = xL1(O2 - O1)

collect like terms,

L2 = xL1(O2 - O1) + L1

factorize the right hand side;

L2 = L1[x(O2 - O1)]

L1 = \frac{L2}{X(O2 -O1)}

2. To make O2 the subject of formula;

x = \frac{L2 - L1}{L1(O2 -O1)}

cross multiply to have;

L2 - L1 = xL1(O2 - O1)

open the bracket to have;

L2 - L1 = xL1O2 - xL1O1

⇒ xL1O2 = L2 - L1 + xL1O1

O2 = \frac{L2-L1 + XL1O1}{XL1}

     = \frac{L2 - L1(1 + XO1)}{XL1}

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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

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3 years ago
A cube made out of lead has a mass of 12 grams. A side of the cube has length 1 cm. What would be the mass of the cube whose sid
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Answer:

<h3><em><u>The mass of a lead cube whose </u></em><em><u>side </u></em><em><u>have length 1.5 cm is 40.5 </u></em><em><u>grams</u></em></h3>

Step-by-step explanation:

<u>1. Let's review the information given to</u>

<u>1. Let's review the information given tous to answer the question correctly:</u>

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Side of the cube <em>= 1 cm</em>

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<em><u>What would be the mass of a leadcube whose sides have length 1.5 cm?</u></em>

<em><u>What would be the mass of a leadcube whose sides have length 1.5 cm?For answering the question, let's</u></em>

<em><u>What would be the mass of a leadcube whose sides have length 1.5 cm?For answering the question, let'scalculate the volume of the cube and</u></em>

<em><u>What would be the mass of a leadcube whose sides have length 1.5 cm?For answering the question, let'scalculate the volume of the cube andthen its density, this way:</u></em>

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<em><u>Now, let's calculate the volume of the</u></em>

<em><u>Now, let's calculate the volume of thesecond cube:</u></em>

<h3><em>Volume = Height Width Length</em></h3><h3><em>Volume = Height Width LengthVolume 1.5 * 1.5 1.5</em></h3><h3><em>Volume = Height Width LengthVolume 1.5 * 1.5 1.5Volume =3.375 cm3</em></h3>

<u>Finally, let's calculate the mass, using</u>

<u>Finally, let's calculate the mass, usingthe density of the lead we already</u>

<u>Finally, let's calculate the mass, usingthe density of the lead we alreadyknow:</u>

<h3>Density = Mass/Volume</h3><h3><em>12 Mass/3.375</em></h3><h3><em>12 Mass/3.375Mass 12 *3.375</em></h3><h3><em>12 Mass/3.375Mass 12 *3.375Mass = 40.5 gramss</em></h3>
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