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vesna_86 [32]
3 years ago
13

Really hard question dont bother.

Mathematics
1 answer:
WITCHER [35]3 years ago
7 0

Answer:

Yessir

Step-by-step explanation:

Friend 8028966, is very smart because ballons are very smart

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When there is no equal sign, the polynomial is called an ___________.
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The answer is expression
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certain computer virus can damage any file with probability 35%, independently of other files. Suppose this virus enters a folde
Ksivusya [100]

Answer:

0.623 is the probability that between 800 and 850 files get damaged.

Step-by-step explanation:

We are given the following information:

We treat virus can damage computer as a success.

P( virus can damage computer) = 35% = 0.35

The conditions for normal distribution are satisfied.

By normal approximation:

\mu = np = 2400(0.35) = 840\\\sigma = \sqrt{np(1-p)} = \sqrt{2400(0.35)(1-0.35)} = $$23.36

We have to evaluate probability that between 800 and 850 files get damaged.

P(800 \leq x \leq 850) = P(\displaystyle\frac{800 - 840}{23.36} \leq z \leq \displaystyle\frac{850-840}{23.36}) = P(-1.712 \leq z \leq 0)\\\\= P(z \leq 0.428) - P(z < -1.712)\\= 0.666 - 0.043 = 0.623 = 62.3\%

P(800 \leq x \leq 850) = 62.3\%

0.623 is the probability that between 800 and 850 files get damaged.

8 0
3 years ago
Write the numbers in the correct order from least to greatest -4, 2 1/5, -2 3/4, 1/10
Naya [18.7K]
-4, -2 3/4, 1/10, 2 1/5
8 0
3 years ago
Solve the system of linear equations by graphing. y = –5/2 x – 7 x + 2y = 4 What is the solution to the system of linear equatio
Anestetic [448]

Keywords

system; linear equations; graphing; solution

we have

y=-\frac{5}{2}x-7 -------> equation A

x+2y=4

x=4-2y ----> equation B

The <u>system</u> of <u>linear equations</u> is composed of equation A and equation B

Substitute the equation B in equation A

y=-\frac{5}{2}[4-2y]-7

y=-10+5y-7

4y=17

y=17/4=4.25

Find the value of x

x=4-2[17/4]=-4.5

The<u> solution</u> is the point (-4.5,4.25)

Using a <u>graphing</u> tool

see the attached figure

therefore

the answer is the option A

(-4.5,4.25)


4 0
3 years ago
Read 2 more answers
EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 11z on the curve of intersection of the plane x − y + z =
Taya2010 [7]

Answer:

\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}

<em>Maximum value of f=2.41</em>

Step-by-step explanation:

<u>Lagrange Multipliers</u>

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

\bigtriangledown  f=\lambda \bigtriangledown  g

for some scalar \lambda called the Lagrange multiplier.

For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is

\bigtriangledown  f=\lambda \bigtriangledown  g+\mu \bigtriangledown  h

The gradient of f is

\bigtriangledown  f=

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in x,y,z,\lambda,\mu.

We have

f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0

Let's compute the partial derivatives

f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0

The Lagrange condition leads to

1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)

Operating and simplifying

1=\lambda+2x\mu\\2=-\lambda +2y\mu \\\lambda=11

Replacing the value of \lambda in the two first equations, we get

1=11+2x\mu\\2=-11 +2y\mu

From the first equation

\displaystyle 2\mu=\frac{-10}{x}

Replacing into the second

\displaystyle 13=y\frac{-10}{x}

Or, equivalently

13x=-10y

Squaring

169x^2=100y^2

To solve, we use the restriction h

x^2 + y^2 = 1

Multiplying by 100

100x^2 + 100y^2 = 100

Replacing the above condition

100x^2 + 169x^2 = 100

Solving for x

\displaystyle x=\pm \frac{10}{\sqrt{269}}

We compute the values of y by solving

13x=-10y

\displaystyle y=-\frac{13x}{10}

For

\displaystyle x= \frac{10}{\sqrt{269}}

\displaystyle y= -\frac{13}{\sqrt{269}}

And for

\displaystyle x= -\frac{10}{\sqrt{269}}

\displaystyle y= \frac{13}{\sqrt{269}}

Finally, we get z using the other restriction

x - y + z = 1

Or:

z = 1-x+y

The first solution yields to

\displaystyle z = 1-\frac{10}{\sqrt{269}}-\frac{13}{\sqrt{269}}

\displaystyle z = \frac{-23\sqrt{269}+269}{269}

And the second solution gives us

\displaystyle z = 1+\frac{10}{\sqrt{269}}+\frac{13}{\sqrt{269}}

\displaystyle z = \frac{23\sqrt{269}+269}{269}

Complete first solution:

\displaystyle x= \frac{10}{\sqrt{269}}\\\\\displaystyle y= -\frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{-23\sqrt{269}+269}{269}

Replacing into f, we get

f(x,y,z)=-0.4

Complete second solution:

\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}

Replacing into f, we get

f(x,y,z)=2.4

The second solution maximizes f to 2.4

5 0
3 years ago
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