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bixtya [17]
3 years ago
12

It’s linear so you map out the first differences and then put it into y=Mx+b form. For the you gotta find the y intercept

Mathematics
1 answer:
Alika [10]3 years ago
6 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Let's find the slope (m) :

  • \dfrac{y2 - y1}{x2 - x1}

  • \dfrac{16 - 11}{9 - 6}

  • \dfrac{5}{3}

slope of given line is 5/3

now, let's plug the value of coordinates of a point lying on the line ( for example : (6 , 11) and slope (m) to find the y - intercept (c) :

  • y = mx + c

  • 11 =   \bigg(\dfrac{5}{ 3}  \times 6 \bigg)  + c

  • 11 = (5 \times 2) + c

  • 11 = 10 + c

  • c = 11 - 10

  • c = 1

value of y - intercept (c) = 1, so our equation will be :

  • y = mx + c

  • y =  \dfrac{5}{3}x  + 1
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P(k)=a^k=2 3 4 find value of a that makes this is a valid probability distribution
Vesna [10]
Sounds like you're asked to find a such that

\displaystyle\sum_{k=2}^4\mathbb P(k)=\mathbb P(2)+\mathbb P(3)+\mathbb P(4)=1

In other words, find a that satisfies

a^2+a^3+a^4=1

We can factorize this as

a^4+a^3+a^2-1=a^3(a+1)+(a-1)(a+1)=(a+1)(a^3+a-1)=0

In order that \mathbb P(k) describes a probability distribution, require that \mathbb P(k)\ge0 for all k, which means we can ignore the possibility of a=-1.

Let a=y+\dfrac xy.

a^3+a-1=\left(y+\dfrac xy\right)^3+\left(y+\dfrac xy\right)-1=0
\left(y^3+3xy+\dfrac{3x^2}y+\dfrac{x^3}{y^3}\right)+\left(y+\dfrac xy\right)-1=0

Multiply both sides by y^3.

y^6+3xy^4+3x^2y^2+x^3+y^4+xy^2-y^3=0

We want to find x\neq0 that removes the quartic and quadratic terms from the equation, i.e.

\begin{cases}3x+1=0\\3x^2+x=0\end{cases}\implies x=-\dfrac13

so the cubic above transforms to

y^6-y^3-\dfrac1{27}=0

Substitute y^3=z and we get

z^2-z-\dfrac1{27}=0\implies z=\dfrac{9+\sqrt{93}}{18}
\implies y=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}
\implies a=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}-\dfrac13\sqrt[3]{\dfrac{18}{9+\sqrt{93}}}
6 0
4 years ago
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