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3 years ago

The graph of a system of inequalities is shown.

Mathematics

2 answers:

dangina [55]3 years ago

8 0

Answer:

The system is

$x+2y \leq -8$

$y> 2x$

Step-by-step explanation:

Part 1

Find the equation of the first inequality

we know that

The first line is a solid line with negative slope passing through the points (-6,-1) and (0,-4)

The slope is equal to

$m=(-4+1)/(0+6)\\m=-0.5$

The equation of the solid line in slope intercept form is

$y=-0.5x-4$

Everything below the line is shaded

The inequality is

$y \leq -0.5x-4$

Convert to standard form

Adds 0.5x both sides

$0.5x+y \leq -4$

Multiply by 2 both sides

$x+2y \leq -8$ -----> First inequality

Part 2

Find the equation of the second inequality

we know that

The second line is a dashed line with positive slope passing through the points (-2,-4) and (0,0)

This line represent a proportional relationship, because the line passes through the origin

The slope is equal to the constant of proportionality

$k=(-4)/(-2)=2$

The equation of the dashed line is

$y=2x$

Everything to the left of the line is shade

The inequality is

$y> 2x$ -----> Second inequality

see the attached figure to better understand the problem

Anni [7]3 years ago

6 0

Answer:

the correct anwser is A.......

y > 2x

x + 2y ≤ –8 just took the unit test

Step-by-step explanation:

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A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

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(a) The probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) n = 28.09

Step-by-step explanation:

The random variable Yₙ is defined as the total numbers of dollars paid in n years.

It is provided that Yₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of Yₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For n = 20 the mean and standard deviation of Y₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that Y₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a z table for probability.

Thus, the probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (Yₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of z for which P (Z < z) = 0.01 is 2.33.

Compute the value of n as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$

The last equation is a quadratic equation.

The roots of a quadratic equation are:

$n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of n = 28.09.

8 0

3 years ago