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Ainat [17]
3 years ago
10

In large population 51% people vaccinated if 4 people are randomly selected what’s the probability at least 1 has been vaccinate

d
Mathematics
1 answer:
OlgaM077 [116]3 years ago
8 0

Answer:

hgktycmjgk,uy

Step-by-step explanation:

hky

You might be interested in
Use Gaussian Elimination to find an equation of a polynomial that passes through points A(-5,-3), B(-2,3). C(3,3), D(6,19). Indi
Marrrta [24]

Answer:

The polynomial equation that passes through the points is 2-\frac{2}{3}x+\frac{1}{12}x^{2}+\frac{1}{12}x^{3}

Step-by-step explanation:

Suppose you have a function y = f(x) which goes through these points

A(-5,-3), B(-2,3). C(3,3), D(6,19)

there is a polynomial P(x) of degree 3 which goes through these point.

We use the fact that <em>four distinct points will determine a cubic function.</em>

P(x) is the degree 3 polynomial through the 4 points, a standard way to write it is

P(x) = a+bx+cx^2+dx^3

Next replace the given points one by one, which leads to a system of 4 equations and 4 variables (namely a,b,c,d)

-3=a+b\cdot-5+c\cdot -5^2+d\cdot -5^3\\3=a+b\cdot-2+c\cdot -2^2+d\cdot -2^3\\3=a+b\cdot 3+c\cdot 3^2+d\cdot 3^3\\19=a+b\cdot 6+c\cdot 6^2+d\cdot 6^3

We can rewrite this system as follows:

-3=a-5\cdot b+25\cdot c-125\cdot d\\3=a-2\cdot b+4\cdot c-8\cdot d\\3=a+3\cdot b+9\cdot c+27\cdot d\\19=a+6\cdot b+36\cdot c+216\cdot d

To use the Gaussian Elimination we need to express the system of linear equations in matrix form (<em>the matrix equation Ax=b</em>).

The coefficient matrix (A) for the above system is

\left[\begin{array}{cccc}1&-5&25&-125\\1&-2&4&-8\\1&3&9&27\\1&6&36&216\end{array}\right]

the variable matrix (x) is

\left[\begin{array}{c}a&b&c&d\end{array}\right]

and the constant matrix (b) is

\left[\begin{array}{c}-3&3&3&19\end{array}\right]

We also need the augmented matrix, it is obtained by appending the columns of the coefficient matrix and the constant matrix.

\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\1&-2&4&-8&3\\1&3&9&27&3\\1&6&36&216&19\end{array}\right]

To transform the augmented matrix to the reduced row echelon form we need to follow these steps:

  • Subtract row 1 from row 2 \left(R_2=R_2-R_1\right)

\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\1&3&9&27&3\\1&6&36&216&19\end{array}\right]

  • Subtract row 1 from row 3 \left(R_3=R_3-R_1\right)

\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\0&8&-16&152&6\\1&6&36&216&19\end{array}\right]

  • Subtract row 1 from row 4 \left(R_4=R_4-R_1\right)

\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&3&-21&117&6\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right]

  • Divide row 2 by 3 \left(R_2=\frac{R_2}{3}\right)

\left[\begin{array}{cccc|c}1&-5&25&-125&-3\\0&1&-7&39&2\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right]

  • Add row 2 multiplied by 5 to row 1 \left(R_1=R_1+\left(5\right)R_2\right)

\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&8&-16&152&6\\0&11&11&341&22\end{array}\right]

  • Subtract row 2 multiplied by 8 from row 3 \left(R_3=R_3-\left(8\right)R_2\right)

\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&40&-160&-10\\0&11&11&341&22\end{array}\right]

  • Subtract row 2 multiplied by 11 from row 4 \left(R_4=R_4-\left(11\right)R_2\right)

\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&40&-160&-10\\0&0&88&-88&0\end{array}\right]

  • Divide row 3 by 40 \left(R_3=\frac{R_3}{40}\right)

\left[\begin{array}{cccc|c}1&0&-10&-70&7\\0&1&-7&39&2\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right]

  • Add row 3 multiplied by 10 to row 1 \left(R_1=R_1+\left(10\right)R_3\right)

\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&-7&39&2\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right]

  • Add row 3 multiplied by 7 to row 2 \left(R_2=R_2+\left(7\right)R_3\right)

\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&88&-88&0\end{array}\right]

  • Subtract row 3 multiplied by 88 from row 4 \left(R_4=R_4-\left(88\right)R_3\right)

\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&264&22\end{array}\right]

  • Divide row 4 by 264 \left(R_4=\frac{R_4}{264}\right)

\left[\begin{array}{cccc|c}1&0&0&30&9/2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right]

  • Subtract row 4 multiplied by 30 from row 1 \left(R_1=R_1-\left(30\right)R_4\right)

\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&11&1/4\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right]

  • Subtract row 4 multiplied by 11 from row 2 \left(R_2=R_2-\left(11\right)R_4\right)

\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&0&-2/3\\0&0&1&-4&-1/4\\0&0&0&1&1/12\end{array}\right]

  • Add row 4 multiplied by 4 to row 3 \left(R_3=R_3+\left(4\right)R_4\right)

\left[\begin{array}{cccc|c}1&0&0&0&2\\0&1&0&0&-2/3\\0&0&1&0&1/12\\0&0&0&1&1/12\end{array}\right]

From the reduced row-echelon form the solutions are:

\left[\begin{array}{c}a=2&b=-2/3&c=1/12&d=1/12\end{array}\right]

The polynomial P(x) is:

2-\frac{2}{3}x+\frac{1}{12}x^{2}+\frac{1}{12}x^{3}

We can check our solution plotting the polynomial and checking that it passes through the points.

3 0
3 years ago
Write in point slope form the equation of a line that has the slope of -4 and the passes
densk [106]

Answer:

point slope form is below. since you have the slope already, all you have to do is plug in the point for x1 and y1 and then the slope for m

y - y1 = m(x -x1)

y - 2 = -4 (x - 2) (this is the equation)

Step-by-step explanation:

4 0
3 years ago
CAN SOMEONE EXPLAIN AND HELP ME ANSWER THIS QUESTION
Rasek [7]
I believe it has a maximum as it’s value
6 0
3 years ago
Are the graphs of y= 1/3x-4 and y-2=-3(x-4) parallel, perpendicular, or neither
melamori03 [73]

Answer:

ghtrh

Step-by-step explanation:

3 0
3 years ago
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Use the x-intercept method to find all real solutions of the equation.<br> x^3-8x^2+9x+18=0
meriva

Answer:

a. x=-1,3,\:or\:6

Step-by-step explanation:

The given equation is;

x^3-8x^2+9x+18=0

To solve by the x-intercept method we need to graph the corresponding function using a graphing calculator or software.

The corresponding function is

f(x)=x^3-8x^2+9x+18

The solution to x^3-8x^2+9x+18=0 is where the graph touches the x-axis.

We can see from the graph  that; the x-intercepts are;

(-1,0),(3,0) and (6,0).

Therefore the real solutions are:

x=-1,3,\:or\:6

8 0
3 years ago
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