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Marysya12 [62]
3 years ago
5

The general process of gathering, organizing, summarizing, analyzing, and interpreting data is called ________. Inferential stat

istics Statistics Descriptive statistics Levels of measurement
Mathematics
2 answers:
Alexus [3.1K]3 years ago
8 0
Idk main math is pretty hard u know
Galina-37 [17]3 years ago
6 0

Answer:

statistics

Step-by-step explanation:

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Whats the answer to: 3x + 5y = 2 9x + 11y = 14
valkas [14]

\begin{gathered}\{\begin{array}{ccc}3x+5y=2&|\cdot(-3)\\9x+11y=14\end{array}\\\underline{+\{\begin{array}{ccc}-9x-15y=-6\\9x+11y=14\end{array}}\ \ |\text{add both sides of equations}\\.\ \ \ \ \ -4y=8\ \ \ |:(=4)\\.\ \ \ \ \ y=-2\\\\\text{substitute the value of y to the first equation}\\\\3x+5\cdot(-2)=2\\3x-10=2\ \ \ |+10\\3x=12\ \ \ |:3\\x=4\\\\Answer:\ x=4;\ y=-2\to(4;\ -2)\end{gathered}

{

3x+5y=2

9x+11y=14

∣⋅(−3)

−9x−15y=−6

9x+11y=14

add both sides of equations

. −4y=8 ∣:(=4)

. y=−2

substitute the value of y to the first equation

3x+5⋅(−2)=2

3x−10=2 ∣+10

3x=12 ∣:3

x=4

Answer: x=4; y=−2→(4; −2)

5 0
3 years ago
Which of the following represents the most accurate estimation of 4+8
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Classify the function as linear or quadratic and identify the quadratic, linear, and constant terms. y=(3x+4)(-2x-3)
MissTica
Y=(3x+4)(-2x-3)=-6x^2-4x-9x-12=-6x^2-13x-12.
It is a quadratic function because it has x in power 2 (x^2)
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The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.
Taya2010 [7]
Given a solution y_1(x)=\ln x, we can attempt to find a solution of the form y_2(x)=v(x)y_1(x). We have derivatives

y_2=v\ln x
{y_2}'=v'\ln x+\dfrac vx
{y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}

Substituting into the ODE, we get

v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0
v''x\ln x+(2+\ln x)v'=0

Setting w=v', we end up with the linear ODE

w'x\ln x+(2+\ln x)w=0

Multiplying both sides by \ln x, we have

w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0

and noting that

\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x

we can write the ODE as

\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0

Integrating both sides with respect to x, we get

wx(\ln x)^2=C_1
w=\dfrac{C_1}{x(\ln x)^2}

Now solve for v:

v'=\dfrac{C_1}{x(\ln x)^2}
v=-\dfrac{C_1}{\ln x}+C_2

So you have

y_2=v\ln x=-C_1+C_2\ln x

and given that y_1=\ln x, the second term in y_2 is already taken into account in the solution set, which means that y_2=1, i.e. any constant solution is in the solution set.
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3 years ago
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