heureka

A customer ordered 15 pieces of gourmet chocolate.
The order can be packaged in small boxes that contain 1, 2 or 4 pieces of chocolate.
Any box that is used must be full.
How many different combinations of boxes can be used for the customer's 15 chocolate pieces?
One such combination to be included is to use seven 2-piece boxes and one 1-piece box.

\( \small{\text{In $15$ pieces max. $\color{red}3$ ($\color{green}4$-piece boxes), max. $\color{red}7$ ($\color{green}2$-piece boxes), and max. $\color{red}15$ ($\color{green}1$-piece boxes) }} \\\\ \)

\(\small{ \begin{array}{|rcll|} \hline && \displaystyle \left(\sum\limits_{i=0}^{\color{red}3} x^{({\color{green}{4}}*i)} \right) \times \left(\sum\limits_{i=0}^{\color{red}7} x^{({\color{green}{2}}*i)} \right) \times \left(\sum\limits_{i=0}^{\color{red}15} x^{({\color{green}{1}}*i)} \right) \\\\ &=&(1+x^4+x^8+x^{12}) \times (1+x^2+x^4+x^6+x^8+x^{10}+x^{12}+x^{14}) \times \\ && \times (1+x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}+x^{13}+x^{14}+x^{15} ) \\\\ &=& x^{41} + x^{40} + 2 x^{39} + 2 x^{38} + 4 x^{37} + 4 x^{36} + 6 x^{35} + 6 x^{34} + 9 x^{33} + 9 x^{32} + 12 x^{31} + 12 x^{30} \\ &+& 16 x^{29} + 16 x^{28} + 20 x^{27} + 20 x^{26} + 22 x^{25} + 22 x^{24} + 24 x^{23} + 24 x^{22} + 24 x^{21} \\ &+& 24 x^{20} + 24 x^{19} + 24 x^{18} + 22 x^{17} + 22 x^{16} + {\color{red}20 x^{15}} + 20 x^{14} + 16 x^{13} + 16 x^{12} \\ &+& 12 x^{11} + 12 x^{10} + 9 x^9 + 9 x^8 + 6 x^7 + 6 x^6 + 4 x^5 + 4 x^4 + 2 x^3 + 2 x^2 + x + 1 \\ \hline \end{array} } \)

\(\text{The coefficient from $\small{ \color[rgb]{1,0,0}{x^{15}} } $ is $\small{ \color[rgb]{1,0,0}{ 20 } } $. } \\ \text{So there are $\mathbf{20}$ possibilities.} \)

\(\begin{array}{|r|r|r|r|} \hline & \color{green}{1}\text{-piece boxes} & \color{green}{2}\text{-piece boxes} & \color{green}{4}\text{-piece boxes} \\ \hline 1 & 15 & & \\ \hline 2 & 13 & 1 & \\ \hline 3 & 11 & 2 & \\ \hline 4 & 11 & & 1 \\ \hline 5 & 9 & 1 & 1 \\ \hline 6 & 9 & 3 & \\ \hline 7 & 7 & & 2 \\ \hline 8 & 7 & 2 & 1 \\ \hline 9 & 7 & 4 & \\ \hline 10 & 5 & 1 & 2 \\ \hline 11 & 5 & 3 & 1 \\ \hline 12 & 5 & 5 & \\ \hline 13 & 3 & & 3 \\ \hline 14 & 3 & 2 & 2 \\ \hline 15 & 3 & 4 & 1 \\ \hline 16 & 3 & 6 & \\ \hline 17 & 1 & 1 & 3 \\ \hline 18 & 1 & 3 & 2 \\ \hline 19 & 1 & 5 & 1 \\ \hline 20 & 1 & 7 & \\ \hline \end{array}\)

Find all real x where

\(\displaystyle 2\cdot \left(\frac{x-5}{x-3}\right) > \frac{2x-5}{x+2} + 5.\)
2\cdot\frac{x-5}{x-3} > \frac{2x-5}{x+2} + 5.

1. rearrange:

\(\begin{array}{|rcll|} \hline 2\cdot \left( \dfrac{x-5}{x-3} \right) &>& \dfrac{2x-5}{x+2} + 5 \\\\ \dfrac{2x-10}{x-3} &>& \dfrac{2x-5+5(x+2)}{x+2} \\\\ \dfrac{2x-10}{x-3} &>& \dfrac{2x-5+5x+10}{x+2} \\\\ \mathbf{ \dfrac{2x-10}{x-3} } & \mathbf{>} & \mathbf{\dfrac{7x+5}{x+2}} \\ \hline \end{array}\)

2.  reduce / convert fractions to a common :

\(\begin{array}{|rcll|} \hline \dfrac{2x-10}{x-3} - \dfrac{7x+5}{x+2} & > & 0 \\\\ \dfrac{(2x-10)(x+2)-(7x+5)(x-3)}{(x-3)(x+2)} & > & 0 \\\\ \dfrac{2x^2+4x-10x-20-7x^2+21x-5x+15}{(x-3)(x+2)} & > & 0 \\\\ \dfrac{-5x^2+10x-5}{(x-3)(x+2)} & > & 0 \\\\ \dfrac{-5(x^2-2x+1)}{(x-3)(x+2)} & > & 0 \quad | \quad : (-5) \\ && \text{ attention } ">" \rightarrow "<" \\\\ \dfrac{x^2-2x+1}{(x-3)(x+2)} & < & 0 \quad | \quad x^2-2x+1 = (x-1)^2 \\\\ \dfrac{(x-1)^2 }{(x-3)(x+2)} & < & 0 \quad | \quad \cdot (x-3)^2(x+2)^2 \\\\ \dfrac{(x-1)^2(x-3)^2(x+2)^2 }{(x-3)(x+2)} & < & 0 \\\\ \mathbf{(x-1)^2(x-3)(x+2)} & \mathbf{\color{red}<} & \mathbf{ 0 } \\ \hline \end{array}\)

3. solve the equation:

\(\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{sign} & (-\infty,-2 ) & -2 & (-2,1) & 1 & (1,3) & 3 & (3,\infty) \\ \hline x-3 & - & - & - & - & - & 0 & + \\ \hline (x-1)^2 & + & + & + & 0 & + & + & + \\ \hline x+2 & - & 0 & + & + & + & + & + \\ \hline \\ \hline \text{sign of } (x-1)^2(x-3)(x+2) & + & 0 & \color{red}- & 0 & \color{red}- & 0 & + \\ \hline \end{array}\)

interval notation: \(\mathbf{(-2,1) \text{ and } (1,3)}\)

Enter (A,B,C,D) in order below if A, B, C, and D are the coefficients of the partial fractions expansion of

\(\displaystyle12\cdot\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2}\)

12\cdot\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2}

\(\small{ \begin{array}{|rcll|} \hline 12\cdot\dfrac{x^3+4}{(x^2-1)(x^2+3x+2)} &=& 12\cdot\dfrac{x^3+4}{(x-1)(x+1)(x+2)(x+1)} \\ \\ \hline \\ 12\cdot\frac{x^3+4}{(x-1)(x+1)(x+2)(x+1)} &=& \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2} \qquad | \qquad \cdot (x-1)(x+2)(x+1)^2 \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline 12(x^3+4) &=& A\cdot (x+2)(x+1)^2 + B\cdot (x-1)(x+2)^2 \\ &+& C\cdot (x-1)(x+2)(x+1) + D\cdot (x-1)(x+2) \\ \hline \end{array}\)

\(\small{ \begin{array}{lrcll} \hline \mathbf{x = 1:} & 12(1 + 4) &=& A\cdot (1+2)(1+1)^2 + B\cdot 0 + C\cdot 0 + D\cdot 0 \\ & 60 &=& A\cdot 3 \cdot 4 \\ & 60 &=& 12A \\ & \mathbf{A} &\mathbf{=}& \mathbf{5} \\ \\ \hline \mathbf{x = -1:} & 12(-1 + 4) &=& A\cdot 0 + B\cdot 0 + C\cdot 0 + D\cdot (-1-1)(-1+2) \\ & 36 &=& -2D \\ & \mathbf{D} &\mathbf{=}& \mathbf{-18} \\ \\ \hline \mathbf{x = -2:} & 12(-8 + 4) &=& A\cdot 0 + B\cdot (-2-1)(-2+1)^2 + C\cdot 0 + D\cdot 0 \\ & -48 &=& -3B \\ & 48 &=& 3B \\ & \mathbf{B} &\mathbf{=}& \mathbf{16} \\ \\ \hline \mathbf{x =0:} & 12(0 + 4) &=& A\cdot 2 + B\cdot (0-1)(0+1)^2 + C\cdot (0-1)(0+2)(0+1)\\ & &+& D\cdot (0-1)(0+2) \\ & 48 &=& 2A-B-2C-2D \\ & 48 &=& 2\cdot 5-16-2C-2\cdot(-18) \\ & 48 &=& 10-16-2C+36 \\ & 48 &=& 30-2C \\ & 2C &=& 30-48 \\ & 2C &=& -18 \\ & \mathbf{C} &\mathbf{=}& \mathbf{-9} \\ \hline \end{array} }\)

\(\displaystyle 12\cdot\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{5}{x-1} + \frac{16}{x+2} - \frac{9}{x+1} - \frac{18}{(x+1)^2}\)

\(\mathbf{(A,B,C,D) = (5,16,-9,-18)}\)

 Enter (A,B,C,D,E) in order below if A, B, C, D, and E are the coefficients of the partial fractions expansion of
\(f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = Ax^2 + Bx + C + \frac{D}{x+2} + \frac{E}{x-1}.\)

 Enter (A,B,C,D,E) in order below if A, B, C, D, and E are the coefficients of the partial fractions expansion of

\(\displaystyle f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = Ax^2 + Bx + C + \frac{D}{x+2} + \frac{E}{x-1}. \)

f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = Ax^2 + Bx + C + \frac{D}{x+2} + \frac{E}{x-1}.

\(\small{ \begin{array}{|rcll|} \hline 3\cdot\dfrac{x^4+x^3+x^2+1}{x^2+x-2} &=& 3\cdot\dfrac{x^4+x^3+x^2+1}{(x+2)(x-1)} \\ \\ \hline \\ 3\cdot\dfrac{x^4+x^3+x^2+1}{(x+2)(x-1)} &=& Ax^2 + Bx + C + \dfrac{D}{x+2} + \dfrac{E}{x-1} \qquad | \qquad \cdot (x+2)(x-1) \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline 3(x^4+x^3+x^2+1) &=& A\cdot x^2(x+2)(x-1) + B\cdot x(x+2)(x-1) \\ &+& C\cdot (x+2)(x-1) + D\cdot (x-1) + E\cdot(x+2) \\ \hline \end{array}\)

\(\small{ \begin{array}{lrcll} \hline \mathbf{x = 1:} & 3(1 +1+1+1) &=& A\cdot 0 + B\cdot 0 + C\cdot 0 + D\cdot 0 + E\cdot (1+2) \\ & 12 &=& 3E \\ & \mathbf{E} &\mathbf{=}& \mathbf{4} \\ \\ \hline \mathbf{x = -2:} & 3(16-8+4+1) &=& A\cdot 0 + B\cdot 0 + C\cdot 0 + D\cdot (-2-1) + E\cdot 0 \\ & 3\cdot 13 &=& -3D \\ & 13 &=& - D \\ & \mathbf{D} &\mathbf{=}& \mathbf{-13} \\ \\ \hline \mathbf{x = 0:} & 3(0+0+0+1) &=& A\cdot 0 + B\cdot 0 + C\cdot(0+2)(0-1) + D\cdot (0-1) + E\cdot (0+2) \\ & 3 &=& -2C-D+2E \\ & 3 &=& -2C-(-13)+2\cdot 4 \\ & 3 &=& -2C+13+8 \\ & 3 &=& -2C+ 21 \\ & 2C &=& 21 -3 \\ & 2C &=& 18 \\ & \mathbf{C} &\mathbf{=}& \mathbf{9} \\ \\ \hline \mathbf{x = -1:} & 3(1 -1+1+1) &=& A\cdot 1(-1+2)(-1-1) + B\cdot (-1)(-1+2)(-1-1) \\ & &+& C\cdot (-1+2)(-1-1) + D\cdot (-1-1) + E\cdot (-1+2) \\ & 6 &=& -2A+2B-2C-2D+E \\ & 6 &=& -2A+2B-2\cdot 9 +2\cdot 13+ 4 \qquad | \qquad : 2 \\ & 3 &=& -A+B-9 +13+ 2 \\ & 3 &=& -A+B+6 \\ & \mathbf{-A+B} &\mathbf{=}& \mathbf{-3} \qquad (1) \\ \\ \hline \mathbf{x = 2:} & 3(16+8+4+1) &=& A\cdot 4(2+2)(2-1) + B\cdot 2(2+2)(2-1) \\ & &+& C\cdot (2+2)(2-1) + D\cdot (2-1) + E\cdot (2+2) \\ & 87 &=& 16A+8B+4C+D+4E \\ & 87 &=& 16A+8B+4\cdot 9 -13+4\cdot 4 \\ & 87 &=& 16A+8B+36 -13+16 \\ & 87 &=& 16A+8B+39 \\ & 16A+8B&=& 87 - 39 \\ & 16A+8B&=& 48 \qquad | \qquad : 8 \\ & \mathbf{2A+B} &\mathbf{=}& \mathbf{6} \qquad (2) \\ \hline \end{array} }\)

\(\begin{array}{|lrclcrclr|} \hline (2) & 2A+B &=& 6 &\qquad & -A+B &=& -3 & (1) \\ & & & &\qquad & B &=& -3+A \\ & 2A-3+A &=& 6 \\ & 3A &=& 9 \\ & \mathbf{A } &\mathbf{=}&\mathbf{ 3} \\ & & & &\qquad & B &=& -3+3 \\ & & & &\qquad & \mathbf{B} &\mathbf{=}&\mathbf{ 0} \\ \hline \end{array} \)

\(\displaystyle f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2} = 3x^2 + 0\cdot x + 9 - \frac{13}{x+2} + \frac{4}{x-1}\)

\(\mathbf{(A,B,C,D,E) = (3,0,9,-13,4)}\)

The following exercise is designed to be solved using technology such as calculators or computer spreadsheets.
If a fair coin is tossed 100 times, the probability that one gets at least 61 heads is given by the following.

This is the p-value if you toss 100 coins and get 61 heads. Use technology to calculate this number. Round your answer to three decimal places.

With calculator texas instrument TI-89:

\(\begin{array}{|rcll|} \hline P(X\ge61) &=& 1 -\text{ binomcdf }(100,0.5,60) \\ &=& 1 - 0.98239989989 \\ &=& 0.01760010011 \\ &\approx& 0.018 \\ \hline \end{array} \)

Find the largest value of x where the plots of 

\(f(x) = - \frac{2x+5}{x+3} \text{ and } g(x) = 4\cdot \frac{x+1}{x-4}\)
intersect.

Find the largest value of x where the plots of

\(\displaystyle f(x) = - \frac{2x+5}{x+3} \qquad \text{ and } \qquad g(x) = 4\cdot \frac{x+1}{x-4}\)

f(x) = - \frac{2x+5}{x+3} \text{ and } g(x) = 4\cdot \frac{x+1}{x-4}
intersect.

\(\begin{array}{|rcll|} \hline - \dfrac{2x+5}{x+3} &=& 4\cdot \frac{x+1}{x-4} \\\\ -(2x+5)(x-4) &=& 4\cdot (x+1)(x+3) \\ 4\cdot (x+1)(x+3) &=& -(2x+5)(x-4) \\ 4\cdot (x^2+3x+x+3) &=& -(2x^2 -8x +5x -20) \\ 4\cdot (x^2+4x+3) &=& -(2x^2 -3x -20) \\ 4x^2+16x+12 &=& -2x^2 +3x +20 \\ 4x^2+2x^2+16x -3x+12-20 &=& 0 \\ 6x^2+13x-8 &=& 0 \\\\ x_{1,2} &=& \dfrac{ -13\pm \sqrt{13^2-4\cdot 6 \cdot (-8) } } {2\cdot 6} \\\\ &=& \dfrac{ -13\pm \sqrt{169+192 } } {12} \\\\ &=& \dfrac{ -13\pm \sqrt{361} } {12} \\\\ &=& \dfrac{ -13\pm 19 } {12} \\\\ x_{\text{max}} &=& \dfrac{ -13+ 19 } {12} \\\\ &=& \dfrac{ 6 } {12} \\\\ &=& \dfrac{ 1 } {2} \\\\ \mathbf{x_{\text{max}}} &\mathbf{=}&\mathbf{ 0.5 } \\ \hline \end{array}\)

Enter (A,B,C) in order below if A, B, and C are the coefficients of the partial fractions expansion of 

\(\frac{2(x^2+x-1)}{x(x^2-1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\)

Enter (A,B,C) in order below if A, B, and C are the coefficients of the partial fractions expansion of 

\(\displaystyle \frac{2(x^2+x-1)}{x(x^2-1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}. \)
 \frac{2(x^2+x-1)}{x(x^2-1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}.

\(\small{ \begin{array}{|rcll|} \hline \dfrac{2(x^2+x-1)}{x(x^2-1)} &=& \dfrac{2(x^2+x-1)}{x(x-1)(x+1)} \\ \\ \hline \\ \dfrac{2(x^2+x-1)}{x(x-1)(x+1)} &=& \dfrac{A}{x} + \dfrac{B}{(x-1)} + \dfrac{C}{(x+1)} \qquad | \qquad \cdot x(x-1)(x+1) \\\\ 2(x^2+x-1) &=& \dfrac{A\cdot x(x-1)(x+1)}{x} + \dfrac{B\cdot x(x-1)(x+1)}{(x-1)} + \dfrac{C\cdot x(x-1)(x+1)}{(x+1)} \\\\ && \boxed{ 2(x^2+x-1) = A (x-1)(x+1) + B x (x+1) + C x(x-1) } \\\\ \hline \end{array} }\)

\(\small{ \begin{array}{lcll} \hline \mathbf{x = 0:} & 2(0+0-1) &=& A (0-1)(0+1) + B \cdot 0 (0+1) + C \cdot 0 (0-1) \\ & -2 &=& -A \\ & \mathbf{A} &\mathbf{=}& \mathbf{2} \\ \\ \hline \mathbf{x = 1:} & 2(1+1-1) &=& A (1-1)(1+1) + B\cdot 1 (1+1) + C \cdot 1(1-1) \\ & 2 &=& A \cdot 0(1+1) + B\cdot 1 (1+1) + C \cdot 1\cdot 0 \\ & 2 &=& 2B \\ & \mathbf{B} &\mathbf{=}& \mathbf{1} \\ \\ \hline \mathbf{x = -1:} & 2(1-1-1) &=& A (-1-1)(-1+1) + B\cdot (-1) (-1+1) + C \cdot (-1)(-1-1) \\ & -2 &=& A (-1-1)\cdot 0 + B\cdot (-1) \cdot 0 + C \cdot (-1)(-1-1) \\ & - 2 &=& 2C \\ & \mathbf{C} &\mathbf{=}& \mathbf{-1} \\ \hline \end{array} }\)

\(\mathbf{(A,B,C) = (2,1,-1)}\)