heureka

Wie hoch ist die Wahrscheinlichkeit, dass von 2 Personen einer die Blutgruppe AB hat

wenn die Gruppe A 43% hat, B 41% hat, 0 11% hat und AB 0,05 hat ?

Wir haben die Wahrscheinlichkeiten:

\(\begin{array}{|lr|r|} \hline \text{Blutgruppe}& A & 43\ \% \\ \text{Blutgruppe}& B & 41\ \% \\ \text{Blutgruppe}& 0 & 11\ \% \\ \text{Blutgruppe}& AB& 0,05 = 5\ \% \\\\ \text{Blutgruppe}& \text{nicht } AB = \overline{AB} & 100\ \% -5\ \% = 95\ \% \\ \hline \end{array}\)

Die Tabelle der Wahrscheinlichkeiten:

P₁ = Person 1

P₂ = Person 2

\(\begin{array}{|l|l|r|r|} \hline && P_2 & P_2 \\ \hline && AB & \overline{AB} \\ \hline P_1 & AB & 5\ \% \cdot 5\ \% & \color{red}{5\ \% \cdot 95\ \% }\\ \hline P_1 & \overline{AB} & \color{red}{95\ \% \cdot 5\ \% } & 95\ \% \cdot 95\ \% \\ \hline \end{array} \)

Die Wahrscheinlichkeit das Person 1 AB hat und die Person 2 kein AB beträgt: \(\color{red}{5\ \% \cdot 95\ \% }\)

Die Wahrscheinlichkeit das Person 2 AB hat und die Person 1 kein AB beträgt: \(\color{red}{95\ \% \cdot 5\ \% }\)

Die Wahrscheinlichkeit, dass von 2 Personen einer die Blutgruppe AB hat beträgt:

\(\begin{array}{|rcll|} \hline && \color{red}{5\ \% \cdot 95\ \% }+\color{red}{95\ \% \cdot 5\ \% } \\ &=& 2\cdot 5\ \% \cdot 95\ \% \\ &=& 2\cdot 0,05 \cdot 95\ \% \\ &=& 0,1 \cdot 95\ \% \\ &=& 9,5\ \% \\ \hline \end{array}\)

bestimmen sie die gleichung der geraden, die durch den punkt P (1/3) geht und parallel zur geraden g(x) =6x+ 4 ist

\(\begin{array}{|rcll|} \hline g(x) &=& 6x+4 \quad \text{Die Steigung der Geraden ist } m = 6\\ \hline \end{array} \)

Die Steigung der parallelen Geraden muss ebenfalls m = 6 sein. Sonst wären die beiden Geraden nicht parallel.

Die paralelle Gerade hat somit erstmal die Gleichung: \(y = 6x + b\)

Ein Punkt auf dieser parallelen Geraden ist P(x=1,y=3)

Wir setzen diesen Punkt in die Gleichung der parallelen Geraden ein und erhalten b.

\(\begin{array}{|rcll|} \hline y &=& 6x + b \\ 3 &=& 6\cdot 1 + b \\ 3 &=& 6+b \\ 3-6 &=& b \\ -3 &=& b \\ \hline \end{array}\)

Somit erhalten wir endlich die parallele Geradengleichung die durch P(x=1,y=3) geht: \(y = 6x - 3\)

Is infinity even, odd or neither?

Infinity is not a real number, it is an idea. An idea of something without an end.

{1, 2, 3, ...} The sequence of natural numbers never ends, and is infinite.

Even a short line segment has infinite points.

what is 39/40-7/20

\(\begin{array}{|rcll|} \hline && \frac{ 39}{40}-\frac{7}{20}\\ &=& \frac{ 39}{2\cdot 20}-\frac{7}{20}\\ &=& \frac{1}{20}\cdot \frac{ 39}{2}-\frac{1}{20}\cdot 7 \\ &=& \frac{1}{20}\cdot \left( \frac{39}{2} - 7 \right) \\ &=& \frac{1}{20}\cdot \left( \frac{39}{2} - \frac{14}{2} \right) \\ &=& \frac{1}{20}\cdot \left( \frac{25}{2} \right) \\ &=& \frac{25}{20\cdot 2} \\ &=& \frac{5\cdot 5}{4\cdot 5 \cdot 2} \\ &=& \frac{5}{4 \cdot 2} \\ &=& \mathbf{\frac{5}{8}} \\ &=& \mathbf{0.625} \\ \hline \end{array} \)

A small lightbulb uses 1 megajoule of energy in a day. What is the wattage of the light bulb?

The megajoule (MJ) is equal to one million (\(10^6\)) joules

One day is equal to \(24\times 60 \times 60 \ \text{seconds} = 86400 \ s\)

Joules to watts calculation:
The power P in watts (W) is equal to the energy E in joules (J),
divided by the time period t in seconds (s):

\(\begin{array}{|rcll|} \hline P(W) &=& \frac{E(J)}{ t(s) } \\ P(W) &=& \frac{10^6\ \text{J}}{ 86400\ \text{s} } \\ P(W) &=& \frac{1000000\ \text{J}}{ 86400\ \text{s} } \\ P(W) &=& \frac{10000}{864}\frac{\text{J}} {\text{s} } \\ P(W) &=& 11.5740740741\frac{\text{J}} {\text{s} } \\ P(W) &=& 11.5740740741\ \text{W}\\ \hline \end{array}\)

(2500[.25(2500/139.931)+ 0.0004(32.8)+0.0001 (4.3722)(1.5))/(1+ 0.0004(32.8)+0.0001 (4.3722)(1.5))

Missing "]"

Which of the following functions has the solution (3, -2)?

Select all that apply.

(1) y=-2x+4

(2) f(x)=-x+2

(3) f(x)=-23x

(4) f(x)=3x−2

(5) f(x)=4x+11

(6) f(x)=6x−20

\(\begin{array}{|rcll|} \hline (3, -2) \quad \text{or } \\ x &=& 3 \quad \text{and }\\ y =f(x) &=& -2 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (1) & y &=& -2x+4 \qquad & | \qquad x = 3 \\ & y &=& -2\times 3 +4 \\ & y &=& -6 +4 \\ & y &=& -2 \qquad \qquad \mathbf{\text{Yes } y = f(x) = -2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (2) & f(x)&=&-x+2 \qquad & | \qquad x = 3 \\ & f(x) &=& -3+2\\ & f(x) &=& -1 \qquad \qquad \text{No } f(x) = -1 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (3) & f(x)&=&-23x \qquad & | \qquad x = 3 \\ & f(x) &=& -23\times 3\\ & f(x) &=& -69 \qquad \qquad \text{No } f(x) = -69 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (4) & f(x)&=&3x-2 \qquad & | \qquad x = 3 \\ & f(x) &=& 3\times 3 -2 \\ & f(x) &=& 9 -2 \\ & f(x) &=& 7 \qquad \qquad \text{No } f(x) = 7 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (5) & f(x)&=&4x+11 \qquad & | \qquad x = 3 \\ & f(x) &=& 4\times 3 +11 \\ & f(x) &=& 12 +11 \\ & f(x) &=& 23 \qquad \qquad \text{No } f(x) = 23 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (6) & f(x)&=&6x-20 \qquad & | \qquad x = 3 \\ & f(x) &=& 6\times 3 -20 \\ & f(x) &=& 18 -20 \\ & f(x) &=& -2 \qquad \qquad \mathbf{\text{Yes } y = f(x) = -2} \\ \hline \end{array}\)

Can someone teach me how to get the inverse of a 3x3 matrix using Gaussian elimination.

Here is an example:

\(\begin{array}{|rcll|} \hline \text{We have Matrix }A = \begin{pmatrix} 1&-2&2\\ 1&0&1\\-1&1&3\end{pmatrix}\\ \hline \end{array}\)

The inverse Matrix \(A^{-1} = \ ?\)

\(\begin{array}{|rcll|} \hline \text{The identity matrix ( unit Matrix ) } I = \begin{pmatrix} 1&0&0\\ 0&1&0\\0&0&1\end{pmatrix}\\ \hline \end{array}\)

Using Gaussian elimination:

\(\left( \begin{matrix} 1&-2&2\\ 1&0&1\\-1&1&3 \end{matrix} \left| \begin{matrix} 1&0&0\\ 0&1&0\\0&0&1 \end{matrix} \right. \right)\)

\( \overset{II-I=II^*}{\curvearrowright} \left( \begin{matrix} 1&-2&2\\ 0&2&-1\\-1&1&3 \end{matrix} \left| \begin{matrix} 1&0&0\\ -1&1&0\\0&0&1 \end{matrix} \right. \right)\)

\( \overset{III+I=III^*}{\curvearrowright} \left( \begin{matrix} 1&-2&2\\ 0&2&-1\\0&-1&5 \end{matrix} \left| \begin{matrix} 1&0&0\\ -1&1&0\\1&0&1 \end{matrix} \right. \right)\)

\( \overset{2\times III+II=III^*}{\curvearrowright} \left( \begin{matrix} 1&-2&2\\ 0&2&-1\\0&0&9 \end{matrix} \left| \begin{matrix} 1&0&0\\ -1&1&0\\1&1&2 \end{matrix} \right. \right)\)

\( \overset{9\times II+III=II^*}{\curvearrowright} \left( \begin{matrix} 1&-2&2\\ 0&18&0\\0&0&9 \end{matrix} \left| \begin{matrix} 1&0&0\\ -8&10&2\\1&1&2 \end{matrix} \right. \right)\)

\( \overset{\frac19 \times II+I=I^*}{\curvearrowright} \left( \begin{matrix} 1&0&2\\ 0&18&0\\0&0&9 \end{matrix} \left| \begin{matrix} \frac19&\frac{10}{9}&\frac{2}{9}\\ -8&10&2\\1&1&2 \end{matrix} \right. \right)\)

\( \overset{ -\frac29 \times III+I=I^*}{\curvearrowright} \left( \begin{matrix} 1&0&0\\ 0&18&0\\0&0&9 \end{matrix} \left| \begin{matrix} -\frac19&\frac{8}{9}&-\frac{2}{9}\\ -8&10&2\\1&1&2 \end{matrix} \right. \right)\)

\( \overset{ \frac{1}{18} \times II=II^*}{\curvearrowright} \left( \begin{matrix} 1&0&0\\ 0&1&0\\0&0&9 \end{matrix} \left| \begin{matrix} -\frac19&\frac{8}{9}&-\frac{2}{9}\\ -\frac{8}{18}&\frac{10}{18}&\frac{2}{18}\\1&1&2 \end{matrix} \right. \right) = \left( \begin{matrix} 1&0&0\\ 0&1&0\\0&0&9 \end{matrix} \left| \begin{matrix} -\frac19&\frac{8}{9}&-\frac{2}{9}\\ -\frac{4}{9}&\frac{5}{9}&\frac{1}{9}\\1&1&2 \end{matrix} \right. \right)\)

\( \overset{ \frac{1}{9} \times III=III^*}{\curvearrowright} \left( \begin{matrix} 1&0&0\\ 0&1&0\\0&0&1 \end{matrix} \left| \begin{matrix} -\frac19&\frac{8}{9}&-\frac{2}{9}\\ -\frac{4}{9}&\frac{5}{9}&\frac{1}{9}\\ \frac{1}{9}&\frac{1}{9}&\frac{2}{9} \end{matrix} \right. \right)\)

\(\begin{array}{|rcll|} \hline \text{The inverse Matrix } A ^{-1} = \begin{pmatrix} -\frac19&\frac{8}{9}&-\frac{2}{9}\\ -\frac{4}{9}&\frac{5}{9}&\frac{1}{9}\\ \frac{1}{9}&\frac{1}{9}&\frac{2}{9} \end{pmatrix} = \frac19 \times \begin{pmatrix} -1&8&-2\\ -4&5&1\\ 1&1&2 \end{pmatrix}\\ \hline \end{array} \)

Aris can paint a room in 5 ½ hours.
His brother can do the same job in 7 ½ hours.
How long will it take them to paint the room together?

Let t = time required when working together

\(\begin{array}{|rcll|} \hline t_1 &=& 5\frac12 \\ t_2 &=& 7\frac12 \\\\ \frac{1}{t} &=& \frac{1}{t_1} + \frac{1}{t_2} \\ \frac{1}{t} &=& \frac{t_1+t_2}{t_1\cdot t_2} \\ \mathbf{t} &\mathbf{=}& \mathbf{ \frac{t_1\cdot t_2}{t_1+t_2} }\\\\ t & = & \frac{t_1\cdot t_2}{t_1+t_2} \qquad & \qquad t_1 = 5\frac12 = \frac{11}{2} \qquad t_2 = 7\frac12 = \frac{15}{2}\\ t & = & \frac{\frac{11}{2}\cdot \frac{15}{2}}{\frac{11}{2}+\frac{15}{2}} \\ t & = & \frac{ \frac{11\cdot 15}{4} } { \frac{11+15}{2} } \\ t & = & \frac{ \frac{165}{4} }{ \frac{26}{2} } \\ t & = & \frac{165}{4} \cdot \frac{2}{26} \\ t & = & \frac{165}{2\cdot26} \\ t & = & \frac{165}{52} \\ t & = & 3.17307692308~ \text{hours}\\ \hline \end{array}\)

It will take them 3.173 hours to paint the room together.

Andre throws 4 dices. How many possibilities are there to throw a total of 6 eyes?

\(\begin{array}{|r|r|} \hline \text{eyes} & \text{possibilities} \\ \hline 1 & 0 \\ 2 & 0 \\ 3 & 0 \\ 4 & 1 \\ 5 & 4 \\ \color{red}6 & \color{red}10 \\ 7 & 20 \\ 8 & 35 \\ 9 & 56 \\ 10 & 80 \\ 11 & 104 \\ 12 & 125 \\ 13 & 140 \\ 14 & 146 \\ 15 & 140 \\ 16 & 125 \\ 17 & 104 \\ 18 & 80 \\ 19 & 56 \\ 20 & 35 \\ 21 & 20 \\ 22 & 10 \\ 23 & 4 \\ 24 & 1 \\ \hline & \text{sum } = 1296\\ \hline \end{array} \)