<img src="https://tex.z-dn.net/?f=4%5Csqrt%7B3%7D%20%2A%202%5Csqrt%7B3%7D" id="TexFormula1" title="4\sqrt{3} * 2\sqrt{3}" alt="4

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LiRa [457]

3 years ago

\sqrt{3} * 2\sqrt{3}" align="absmiddle" class="latex-formula">

Mathematics

2 answers:

Marysya12 [62]3 years ago

5 0

Answer: 24

Step-by-step explanation:

yanalaym [24]3 years ago

4 0

Answer:

12.89

Step-by-step explanation:

use a calculator

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Solve for x: 12-2x=-2(y-x)

a_sh-v [17]

12 - 2x = -2(y - x)

12 - 2x = -2y - (-2x)

12 - 2x = -2y + 2x

12 - 2x - 2x = -2y + 2x - 2x

12 - 4x = -2y

12 - 12 - 4x = -2y - 12

-4x = -2y - 12

-4x/4 = -2y/4 - 12/4

-x = -0.5y - 3

-x/-1 = -0.5y/-1 - 3/-1

x = 0.5y + 3

or

x = 3 + 0.5y

3 0

3 years ago

Find a positive number for which the sum of it and its reciprocal is the smallest (least) possible. Let x be the number and let

Allisa [31]

Answer:

$S(x) = x + \frac{1}{x}$ --- Objective function

Interval = $\{x:x=1\}$

Step-by-step explanation:

Given

Represent the number with x

The required sum can be represented as:

$x + \frac{1}{x}$

Hence, the objective function is:

$S(x) = x + \frac{1}{x}$

To get the the interval, we start by differentiating w.r.t x

<em>Using first principle, this gives:</em>

$S'(x) = 1 - \frac{1}{x^2}$

Equate S'(x) to 0 in order to solve for x

$0 = 1 - \frac{1}{x^2}$

Subtract 1 from both sides

$0 -1 = 1 -1 - \frac{1}{x^2}$

$-1 = - \frac{1}{x^2}$

Multiply both sides by -1

$1 = \frac{1}{x^2}$

Cross Multiply

$x^2 * 1 = 1$

$x^2 = 1$

Take positive square root of both sides because x is positive

$\sqrt{x^2} = \sqrt{1$

$x = 1$

Representing x using interval notation, we have

Interval = $\{x:x=1\}$

To get the smallest sum, we substitute 1 for x in $S(x) = x + \frac{1}{x}$

$S(1) = 1 + \frac{1}{1}$

$S(1) = 1 + 1$

$S(1) = 2$

<em>Hence, the smallest sum is 2</em>

3 0

3 years ago

Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a

Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

$P(X$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$