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3 years ago

Which point would be a solution to the system of linear inequalities ?

Mathematics

1 answer:

brilliants [131]3 years ago

8 0

Answer:

(12,-6)

Step-by-step explanation:

we have

$y\leq \frac{4}{3}x+5$ ----> inequality A

$y\geq -\frac{5}{2}x+5$ ---> inequality B

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)

<u><em>Verify each point</em></u>

Substitute the value of x and the value of y of each ordered pair in the inequality A and in the inequality B

case 1) (0,-1)

Inequality A

$-1\leq \frac{4}{3}(0)+5$

$-1\leq5$ ----> is true

Inequality B

$-1\geq -\frac{5}{2}(0)+5$

$-1\geq 5$ ----> is not true

therefore

The ordered pair is not a solution of the system

case 2) (0,3)

Inequality A

$3\leq \frac{4}{3}(0)+5$

$3\leq5$ ----> is true

Inequality B

$3\geq -\frac{5}{2}(0)+5$

$3\geq 5$ ----> is not true

therefore

The ordered pair is not a solution of the system

case 3) (-6,-6)

Inequality A

$-6\leq \frac{4}{3}(-6)+5$

$-6\leq-3$ ----> is true

Inequality B

$-6\geq -\frac{5}{2}(-6)+5$

$-6\geq 20$ ----> is not true

therefore

The ordered pair is not a solution of the system

case 4) (12,-6)

Inequality A

$-6\leq \frac{4}{3}(12)+5$

$-6\leq21$ ----> is true

Inequality B

$-6\geq -\frac{5}{2}(12)+5$

$-6\geq -25$ ----> is true

therefore

The ordered pair is a solution of the system (makes true both inequalities)

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Occasionally an airline will lose a bag. Suppose a small airline has found it can reasonably model the number of bags lost each

julia-pushkina [17]

Answer:

a) The probability that the airline will lose no bags next monday is 0.1108

b) The probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227

c) I would recommend taking a Poisson model with mean 4.4 instead of a Poisson model with mean 2.2

Step-by-step explanation:

The probability mass function of X, for which we denote the amount of bags lost next monday is given by this formula

$P(X=k) = \frac{e^{-2.2} * {2.2}^k }{k!}$

$P(X=0) = \frac{e^{-2.2} * {2.2}^0 }{0!} = 0.1108$

The probability that the airline will lose no bags next monday is 0.1108.

b) Note that $P(X \in \{0,1,2\} = P(X=0) + P(X=1) + P(X=2)$ . And

$P(X=0)+P(X=1)+P(X=2) = e^{-2.2} * (1 + 2.2 + 2.2^2/2) = 0.6227$

Therefore, the probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227.

c) If the double of flights are taken, then you at least should expect to loose a similar proportion in bags, because you will have more chances for a bag to be lost. WIth this in mind, we can correctly think that the average amount of bags that will be lost each day will double. Thus, i would double the mean of the Poisson model, in other words, i would take a Poisson model with mean 4.4, instead of 2.2.

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3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

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3 years ago

Two consecutive integers have a sum of 513. find the smaller of these two number

nexus9112 [7]

513/2=256.5
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san4es73 [151]

Answer:

h(x) * s(x) = 200(1.05)^(x - 1)

Step-by-step explanation:

Our interest equation is s(x) = (1.05)^(x - 1). This is actually a part of a bigger formula for calculating the amount of money accumulated including interest:

A = P(1 + r)^n, where A is the total, P is the principal amount (initial amount), r is the interest rate, and n is the time

Here, we technically already have the (1 + r)^n part; it's just (1.05)^(x - 1). The principle, though, will actually be the 200 because she starts out at $200.

Thus, to combine these, we simply multiply them together to get:

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