Jawaban Matematika Mid Semester

Nama : Dimas Fadli Saputra
Kelas : 12 TKJ 1

1. P1 : jika saya tidak makan, maka saya sakit
    P2 : saya tidak makan
    Kesimpulan : saya sakit


2. P1 : jika panen berhasil maka kesejahteraan petani meningkat
    P2 : kesejahteraan petani tidak meningkat
    Kesimpulan : panen tidak berhasil

3. P1 : jika 2x2 = 4 maka faktor dari 20
    P2 : jika 4 faktor dari 20 maka 20 bilangan genap
    Kesimpulan : jika 2x2 = 4 maka 20 bilangan genap

4.
$\left( -1,\sqrt{3} \right)$
${{r}^{2}}={{x}^{2}}+{{y}^{2}}$
$${{r}^{2}}=-{{1}^{2}}+{{\left( \sqrt{3} \right)}^{2}}$$
${{r}^{2}}=1+3$
$r=\sqrt{4}=2$

$$\alpha {}^\circ =Arc\tan \frac{\sqrt{3}}{-1}$$
      $=Arc\tan -\sqrt{3}$
      $=-60{}^\circ$
$\left( 2,-60{}^\circ  \right)$

5.  $\left( 4,30{}^\circ  \right)$

$$x=r\cos \alpha {}^\circ$$  

$x=4\cos 30{}^\circ$

$x=4.\frac{1}{2}\sqrt{3}$ 

$x=2\sqrt{3}$ 

$y=r\sin \alpha {}^\circ$ 

$y=4\sin 30{}^\circ$ 

$y=2\left( \frac{1}{2} \right)$

$y=1$ 

$\left( 2\sqrt{3},1 \right)$  


6.  $\sin \left( a+b \right)=\sin a.\cos b+\cos a.\sin b$

$\sin 75=\sin \left( 30+45 \right)$

$=\sin 30.\cos 45+\sin 30.\cos 45$

$$=\frac{1}{2}.\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3}.\frac{1}{2}\sqrt{2}$$

$=\frac{1}{4}\sqrt{2}+\frac{1}{4}\sqrt{6}$

$=\frac{1}{4}\left( \sqrt{2}+\sqrt{6} \right)$ 

7.    $\cos \left( 45{}^\circ +60{}^\circ  \right)=\cos 45{}^\circ +\cos 60{}^\circ$ 

$$=\frac{1}{2}\sqrt{2}+\frac{1}{2}$$

$=\frac{1}{2}\sqrt{2}$ 

8.    NS = $\begin{align}

 & 10C3=\frac{10!}{7!3!}=\frac{10.9.8.7!}{7!.5!}=120 \\

  &  \\

  \end{align}$

            NA =  $\begin{align}

  & 3C2.1C1 \\

 & 3.1=3 \\

\end{align}$ Peluangnya \[\frac{5}{120}=\frac{1}{2}\]

9.       Dadu $5=4$               Kemungkinan $=36$

             Dadu $8=\frac{5}{9}$             Peluang $\frac{9}{36}=\frac{1}{4}$ 

10.   Me = \[\begin{align}

& Tb\frac{\left( \frac{n}{2}-Fk \right)}{Fme}.I \\

\end{align}\]

            = $55,5\frac{\left( \frac{20}{2}-9 \right)3}{3}$

            = $\frac{55,5.10-9.3}{3}=55,5$

            Mo = \[\begin{align}

  & Tb\frac{d1}{d1+d2}.I \\

 &  \\

\end{align}\]

            = $52,5+\frac{1}{1+3}.3$

            = $52,5+\frac{1}{4}.3$

            = $52,5+\frac{3}{4}=60$

      Q1 = ….

      Letak = $\frac{20}{4}=5$

      FQ = 5

      Tb = $52,5$

      Fk = 4

      I = 3

      Q1 = $52,5+\frac{5-4}{5}.3$

            = $52,5+\frac{1}{5}.3$

            = $52,5+0,2.3$

            = $52,5+6,6$

            = $53,1$

      Q3 = ….

      Letak = $\frac{20.3}{4}=15$

      FQ = 6

      Tb = $61,5$

      Fk = $14$

      I = 3

      Q1 = $61,5+\frac{15-4}{6}.3$

            = $61,5+\frac{1}{6}.3$

            = $61,5+0,5$

            = $62$

  

   Mean = $\frac{\sum\limits_{{}}^{{}}{F1.X1}}{\sum\limits_{{}}^{{}}{F1}}$

            = $\frac{1143}{20}$

            = $57,15$

11. $\begin{align}
 & \int{\left( x-2 \right)}\left( x+3 \right)dx= \\

 & \int{\left( {{x}^{2}}+3x-2x-6 \right)dx} \\

 \end{align}$ 

 = $\frac{1}{3}{{x}^{3}}+\frac{1}{2}{{x}^{2}}-6x$ 

12.              $\int\limits_{1}^{2}{\left( 2x-3 \right)dx}$

             = ${{x}^{2}}-3x\int\limits_{1}^{2}{{}}$

             = $\left( {{2}^{2}}-3.2 \right)-\left( {{1}^{2}}-3.1 \right)$

             = $\left( 4-6 \right)-\left( 1-3 \right)$

             = $-2-2=0$ 

13.       y = $9-{{x}^{2}}$

$\int\limits_{-1}^{1}{\left( {{x}^{3}} \right)}dx=\frac{1}{4}.{{x}^{4}}$ 

                     = $\frac{1}{4}{{.1}^{4}}-\frac{1}{4}-{{1}^{4}}$

                     = $\frac{1}{4}--\frac{1}{4}$

                     = $\frac{1}{2}$ satuan