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- Question 3 Exercise 6.4 @math-11-kpk:sol:unit06
- ains eight questions. If a student guesses the answer for each question, find the probability that $8$ answers are correct. ====Solution==== We have $8$ questions, each question has two options. Therefore, The... ual outcome to occur is $$\dfrac{1}{256}$$ $8$ answers are correct. Let $$A=\{8\}$$ Obviously only on
- Question 6, Exercise 1.3 @math-11-kpk:sol:unit01
- =0$$ $$(z^2+z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{-1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac... 2}$$ $$(z^2-z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{... t{3}i}{2}$$ The value of $z$ from both equations, we have $$z=\pm \dfrac{1}{2}\pm \dfrac{\sqrt{3}}{2}i... $z^2-2z+4=0$$ According to the quadratic formula, we have $a=1$, $b=-2$ and $c=4$ Thus, we have \begin
- Question 1 Review Exercise 6 @math-11-kpk:sol:unit06
- correct option. <panel> i. In how many ways can we name the vertices of pentagon using any five of t... ) $4880$ \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">(a): $25... ts $\{1,2,3,4,5,6,7\}$ if repeated digits are allowed? * %%(a)%% $14$ * (b) $42$ * %%(c)... (d) $21$ \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(c): $28
- Question 1 Review Exercise 7 @math-11-kpk:sol:unit07
- correct option. <panel> i. In how many ways can we name the vertices of pentagon using any five of t... ) $4880$ \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">(a): $25... ts $\{1,2,3,4,5,6,7\}$ if repeated digits are allowed? * %%(a)%% $14$ * (b) $42$ * %%(c)... (d) $21$ \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(c): $28
- Question 2 Exercise 4.3 @math-11-kpk:sol:unit04
- OOD ====Solution==== Given: $a_1=2, n=17, d=3$ \\ We need to find $a_{17}$ and $S_{17}$. As we know $$a_{n}=a_1+(n-1)d.$$ Thus $$a_{17}=2+(17-1)(3)=50.$$ ... ution==== Given: $a_1=-40$ and $S_{21}=210$.\\ So we have $n=21$ and we have to find $a_{21}$ and $d$. As \begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \
- Question 7 and 8 Exercise 6.2 @math-11-kpk:sol:unit06
- rom the digits $1,2,3,4$ and 5 if repetitions allowed? ====Solution==== There are three places (hundre... lled by five digits, Moreover repetition is allowed. Hence $E_1$ occurs in $m_1=5$ ways only $E_2$... digits $1,2,3,4$ and 5 if repetitions are not allowed? ====Solution==== If repetition is not allowed then each digit can appear once in each number. In t
- Question 1 and 2 Exercise 6.5 @math-11-kpk:sol:unit06
- ac{1}{2}$.\\ Find $P(A \cap B)$. ====Solution==== We know by addition law of probability \begin{align}... ign} Substituting $P(A), P(B)$ and $P(A \cup B)$, we get $$P(A \cap B)=\dfrac{2}{5}+\dfrac{2}{5}-\dfra... dfrac{3}{4}$. Find $P(A \cap B)$ ====Solution==== We are given: $$P(A)=\dfrac{1}{2}, P(\bar{B})=\dfrac{5}{8}, P(A \cup B)=\dfrac{3}{4}$$ We know by complementary events $$P(B)=1-P(\bar{B})$
- Question 5 and 6 Exercise 7.3 @math-11-kpk:sol:unit07
- omial expansion and neglecting $x^2$ and higher powers of $x$. $-\left(1+\frac{2}{3} \cdot \frac{3 x}{8}+\right.$ higher powers of $\left.x\right) x$ $\left(1-\frac{3 x}{2}+\right.$ higher powers of $\left.x\right) \times$ $\left(1+\frac{1}{2}, \frac{5 x}{4}+\right.$ higher powers of $\left.x\right)$ $-\left(1+\frac{x}{4}\right
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- }$ and $C=\begin{bmatrix}x\\y\\z\end{bmatrix}$.\\ We have to prove that $$(AB)C=A(BC)$$ First, we take \begin{align} AB&=\begin{bmatrix}x & y & z\end{bmat... fyz+c{{z}^{2}} \right] \ldots (1) \end{align} Now we take \begin{align}BC&=\begin{bmatrix}a & h & g\\h... } \right] \ldots (2)\end{align} From (1) and (2), we have $$(AB)C=A(BC).$$ =====Question 3(ii)(a)===
- Question 1 Review Exercise 3 @math-11-kpk:sol:unit03
- (d) $3$ \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">(a): $0$... osceles \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(a): Equ... %% $-2$ \\ <btn type="link" collapse="a3">See Answer</btn><collapse id="a3" collapsed="true">(d): $-2... of these \\ <btn type="link" collapse="a4">See Answer</btn><collapse id="a4" collapsed="true">(b): $\v
- Question 16 Exercise 4.2 @math-11-kpk:sol:unit04
- =Question 16===== Insert five arithmetic means between $5$ and $8$ and show that their sum is five times the arithmetic mean between $5$ and $8$. GOOD ====Solution==== Let $A_1, A_2, A_3, A_4, A_5$ be five arithmetic means between $5$ and $8$. Then $5, A_1, A_2, A_3, A_4, A_5, ... re in A.P, where $$a_1=5 \text{ and } a_7=8.$$ As we have \begin{align}&a_7=a+6d\\ \implies &8=5+6d\\
- Question 1 Exercise 5.3 @math-11-kpk:sol:unit05
- onstants on the both sides of the above equation, we get $$A+B=0 \text{and} A=1$$ Putting $A=1$,then \... 1)}$$ Multiplying both sides by $(2 n-1)(2 n+1)$. we get, \begin{align} & \mathrm{I}=A(2 n+1)+B(2 n-1)... Solving the above two equations for $A$ and $B$, we get \begin{align}A&=\dfrac{1}{2}\\ \text{and} B&=... n+2}$$ Multiplying both sides by $(3 n-1)(3 n+2)$ we get, \begin{align} 1&=A(3 n+2)+B(3 n-1) \\ \Right
- Question 5 and 6 Exercise 6.3 @math-11-kpk:sol:unit06
- {12} C_2=66$ lines. The question now is whether we have counted any line twice. the the answer is "No," because there are no three of the given points ... l points are eight so $n=12$. Now for triangles, we have ${ }^{12} C_3=220$ ways to choose the vertices. Again the question is whether we have counted any triangle twice. Again, the answ
- Question 12 Exercise 7.3 @math-11-kpk:sol:unit07
- \ldots$ then show that $4 y^2+4 y-1=0$. Solution: We are given $$ 2 y=\frac{1}{2^2}+\frac{1.3}{2 !} \c... $$ Adding 1 to both sides of the above equation, we get $S=2 y+1=1+\frac{1}{2^2}+\frac{1.3}{2 !} \cdo... dots \end{aligned} $$ Comparing both the series, we have $n x=\frac{1}{2^2}=\frac{1}{4}.... (1)$ and ... \cdot \frac{1}{2^4} $$ Taking square of Eq.(1), we have $n^2 x^2=\frac{1}{16}$ Dividing Eq.(2) by Eq
- Question 3 & 4, Exercise 3.2 @math-11-kpk:sol:unit03
- Solution==== Given $$\vec{r}=p\vec{a}+q\vec{b}.$$ We put the values of $\vec{r},\vec{a}$ and $\vec{b}$ in the given equation. We get $$\hat{i}-9\hat{j}=p(\hat{i}+2\hat{j})+q(5\ha... paring the coeffients of $\hat{i}$ and $\hat{j}$, we have, $$p+5q=1…(i)$$ $$2p-q=-9 …(ii)$$ Multiply $2$ by (i) and subtract (ii) from (i). We have \[\begin{array}{ccc} 2p&+10q&=2 \\ \mathop