heureka

binompdf (6,.4,2) ?

\(\text{binompdf (6,.4,2) } ?\\ \boxed{~ \text{binompdf } (n,\ p,\ x) =\binom{n}{x} \cdot p^x\cdot q^{n-x} ~}\\ \)

\(\begin{array}{rcl} p&=& 0.4\\\\ q&=&(1-p) \\ &=& 1-0.4\\ &=&0.6\\\\ n &=& 6 \\ x &=& 2 \\\\ \text{binompdf } (6,.4,2)=\\ P(X=2) &=& \binom{6}{2}\cdot 0.4^2\cdot0.6^4 \\ &=& \frac62\cdot\frac51\cdot 0.4^2\cdot0.6^4 \\ &=& 15\cdot 0.4^2\cdot0.6^4 \\ &=& 15\cdot 0.16\cdot0.6^4 \\ &=& 15\cdot 0.16\cdot0.1296 \\ &=&0.31104\\ &=& 31.104\ \% \end{array}\)

At camp today, kids can choose art, swimming, or music. Of those activities, 40 kids choose swimming, 16 choose art, and 24 choose music. What is the ratio of kids who choose art to kids who choose swimming?

kids who choose art : kids who choose swimming = \(\frac{16}{ 40}=\frac{2\cdot 8}{ 5\cdot 8} = \frac25\)

(y^2+5y-7)-(y^2-3y-4)

\(\begin{array}{rcll} (y^2+5y-7)-(y^2-3y-4) &=&\\ &=& y^2+5y-7-(y^2-3y-4)\\ &=& y^2+5y-7-y^2+3y+4\\ &=& y^2-y^2+5y+3y-7+4 \qquad & | \qquad y^2-y^2 = 0\\ &=& 5y+3y-7+4 \qquad & | \qquad 5y+3y= 8y\\ &=& 8y-7+4 \qquad & | \qquad-7+4 =-3\\ &=& 8y-3 \\ \mathbf{(y^2+5y-7)-(y^2-3y-4)} &\mathbf{=}& \mathbf{8y-3 } \end{array}\)

The value from :

\(\begin{array}{rcl} ^2\log{(6^2)} - ^2\log{(3^2)} &=&\\ &=& \log_2{(6^2)}-\log_2{(3^2)} \\ &=& \log_2{ ( \frac{6^2 } {3^2 } ) } \\ &=& \log_2{ ( { ( \frac63 ) }^2 ) } \\ &=& \log_2{ ( { 2 }^2 ) } \qquad = \quad ^2\log{ ( { 2 }^2 ) }\\ &=& 2 \end{array}\)

As far as I can gather ...

was ist eine natürliche zahl ?

Zu den natürlichen Zahlen gehören alle Zahlen, mit deren Hilfe beliebige Objekte gezählt werden können.

Die natürlichen Zahlen ohne Null: Sie umfasst die positiven ganzen Zahlen
|N = {1,2,3,4,5,…}

Die natürlichen Zahlen mit Null: : Sie umfasst die nichtnegativen ganzen Zahlen
\(N_0\) = {0,1,2,3,4,5,…}

-2(5/4-3/4)+5y=17

\(\begin{array}{rcl} -2(\frac54-\frac34)+5y &=& 17 \\ -2(\frac{5-3}{4})+5y &=& 17 \\ -2(\frac24)+5y &=& 17 \\ -\frac44+5y &=& 17 \\ -1+5y &=& 17 \qquad | \qquad +1\\ 5y &=& 18 \qquad | \qquad :5\\ y &=& \frac{18}{5} \\ y &=& 3\frac{3}{5} \\ y &=& 3.6 \end{array}\)

3(17.5+11)* 10e-9/ 4* (5 *10e-9)* (220*10 e-9)^3)^0.5

One "(" missing.

wie kann ich die darstellung der ausgabe in eine wissenschaftliche umstellen? also runden... anstatt 1300000 = 13 * 10^5

Eine Vorkommastelle ( 1...9) setzen, also nur das Komma verschieben :

  \(1300000 = 1,3 \cdot 10^{6}\)

how to simplify (4 - 8x)(4 + 8x)

The right answer is:

\(\begin{array}{rcl} (4 - 8x)(4 + 8x) &=& \\ &=&4^2 - (8x)^2\\ &=& 16 -64\cdot x^2\\\\ \mathbf{(4 - 8x)(4 + 8x) } & \mathbf{=} & \mathbf{16 -64\cdot x^2} \end{array}\)

1-16+246-4096+...+4294967296

If you mean:  1 - 16 + 256 - 4096 +-... + 4294967296          256 instead of 246 !!!

We have:

\(\begin{array}{rcl} a_1&=&1\\ a_2&=&- 16 \\ a_3&=&256 \\ a_4&=&- 4096 \\ \dots \\ a_9 &=& 4294967296\\ \boxed{~ r = \frac{a_{n}}{a_{n-1}} ~}\\ r&=&\frac{ a_4 }{ a_3 }=\frac{ a_3 }{ a_2 }=\frac{ a_2 }{ a_1 }\\ r&=&\frac{ - 4096}{ 256 }=\frac{ 256 }{ - 16 }=\frac{ - 16 }{ 1 }\\ r&=& - 16= - 16= - 16\\ \mathbf{r} &\mathbf{=}& \mathbf{- 16}\\ \boxed{~ a_n = a_1\cdot r^{n-1} \\ a_n = a_1\cdot (-16)^{n-1} ~}\\ a_1 &=& 1\cdot \left( - 16 \right)^{1-1} = 1\\ a_2 &=& 1\cdot \left( - 16 \right)^{2-1} = 1\cdot - 16= - 16\\ a_3 &=& 1\cdot \left( - 16 \right)^{3-1} = 1\cdot \left( - 16 \right)^2 = 256\\ a_4 &=& 1\cdot \left( - 16 \right)^{4-1} = 1\cdot \left( - 16 \right)^3 = - 4096 \\ \dots \\ a_{9} &=& 1\cdot \left(- 16 \right)^{9-1} = 1\cdot \left( - 16 \right)^8 = 4294967296\\ \boxed{~ sum_n = a_1\cdot\frac{1-r^n}{1-r} ~}\\ sum_9 &=& 1 \cdot \frac{1-(-16)^9}{1-(-16)} \\ sum_9 &=&\frac{1+68\ 719\ 476\ 736}{17} \\ sum_9 &=&\frac{68\ 719\ 476\ 737}{17} \\ \mathbf{sum_9} &\mathbf{=}& \mathbf{4\ 042\ 322\ 161}\\ \end{array}\)

1 - 16 + 256 - 4096 +-... + 4294967296 = 4 042 322 161