2019-08-10T11:00:21-04:00
QUESTION 1
Test marks for ten students are 2, 3, 6, 7, 7, x, 5, 5, 8, 9
(i) Given that the average mark is 6, find the value of x.
(ii) Calculate the standard deviation for these set of marks
QUESTION 2
Solve the following pair of equations by Cramer’s rule.
1) 3x + 5y = 2
2) 2x-7y = - 1
1
2019-08-11T09:54:10-0400
1.i)
x + 2 + 3 + 6 + 7 + 7 + 5 + 5 + 8 + 9 = 6 ( 10 ) x+2+3+6+7+7+5+5+8+9=6(10) x + 2 + 3 + 6 + 7 + 7 + 5 + 5 + 8 + 9 = 6 ( 10 )
x + 52 = 60 x+52=60 x + 52 = 60
x = 8 x=8 x = 8 ii)
( 2 − 6 ) 2 + ( 3 − 6 ) 2 + 2 ( 5 − 6 ) 2 + ( 6 − 6 ) 2 + 2 ( 7 − 6 ) 2 + 2 ( 8 − 6 ) 2 + ( 9 − 6 ) 2 = 46 (2-6)^2+(3-6)^2+2(5-6)^2+(6-6)^2+2(7-6)^2+2(8-6)^2+(9-6)^2=46 ( 2 − 6 ) 2 + ( 3 − 6 ) 2 + 2 ( 5 − 6 ) 2 + ( 6 − 6 ) 2 + 2 ( 7 − 6 ) 2 + 2 ( 8 − 6 ) 2 + ( 9 − 6 ) 2 = 46 The standard deviation for these set of marks:
s = 46 9 − 1 = 2.4 s=\sqrt{\frac{46}{9-1}}=2.4 s = 9 − 1 46 = 2.4 2. The coefficient matrix is
[ 3 5 2 − 7 ] \begin{bmatrix}
3 & 5 \\
2 & -7
\end{bmatrix} [ 3 2 5 − 7 ] And
∣ 3 5 2 − 7 ∣ = 3 ( − 7 ) − 2 ( 5 ) = − 31 \begin{vmatrix}
3 & 5 \\
2 & -7
\end{vmatrix}=3(-7)-2(5)=-31 ∣ ∣ 3 2 5 − 7 ∣ ∣ = 3 ( − 7 ) − 2 ( 5 ) = − 31 So,
x = ∣ 2 5 − 1 − 7 ∣ − 31 = ( 2 ) ( − 7 ) − ( − 1 ) ( 5 ) − 31 = 9 31 x=\frac{\begin{vmatrix}
2 & 5 \\
-1 & -7
\end{vmatrix}}{-31}=\frac{(2)(-7)-(-1)(5)}{-31}=\frac{9}{31} x = − 31 ∣ ∣ 2 − 1 5 − 7 ∣ ∣ = − 31 ( 2 ) ( − 7 ) − ( − 1 ) ( 5 ) = 31 9
y = ∣ 3 2 2 − 1 ∣ − 31 = ( 3 ) ( − 1 ) − ( 2 ) ( 2 ) − 31 = 7 31 y=\frac{\begin{vmatrix}
3 & 2 \\
2 & -1
\end{vmatrix}}{-31}=\frac{(3)(-1)-(2)(2)}{-31}=\frac{7}{31} y = − 31 ∣ ∣ 3 2 2 − 1 ∣ ∣ = − 31 ( 3 ) ( − 1 ) − ( 2 ) ( 2 ) = 31 7
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