Answer:
Step-by-step explanation:
½(x-2) = 8 + 5x
Multiply both sides by 2
2⋅½(x-2) = 2(8 + 5x)
1(x-2) = 2(8 + 5x)
Apply the distributive rule
x - 2 = 16 + 10x
Subtract x from both sides
-2 = 16 + 9x
Subtract 16 from both sides
-18 = 9x
Divide both sides by the coefficient of x
-18 ÷ 9 = 9x ÷ 9
-2 = x
Answer: a) critical value = 0.0853, b) we reject the null hypothesis.
Step-by-step explanation:
Since we have given that
z = -1.37
And the hypothesis are given below:
Since α = 0.01
since critical value = 0.0853
As we can see that 0.853 < 0.25.
so, we reject the null hypothesis.
Hence, a) critical value = 0.0853, b) we reject the null hypothesis.
i)On z, define a∗b=a−b
here aϵz
+
and bϵz
+
i.e.,a and b are positive integers
Let a=2,b=5⇒2∗5=2−5=−3
But −3 is not a positive integer
i.e., −3∈
/
z
+
hence,∗ is not a binary operation.
ii)On Q,define a∗b=ab−1
Check commutative
∗ is commutative if,a∗b=b∗a
a∗b=ab+1;a∗b=ab+1=ab+1
Since a∗b=b∗aforalla,bϵQ
∗ is commutative.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(ab+1)∗c=(ab+1)c+1=abc+c+1
a∗(b∗c)=a∗(bc+1)=a(bc+1)+1=abc+a+1
Since (a∗b)∗c
=a∗(b∗c)
∗ is not an associative binary operation.
iii)On Q,define a∗b=
2
ab
Check commutative
∗ is commutative is a∗b=b∗a
a∗b=
2
ab
b∗a=
2
ba
=
2
ab
a∗b=b∗a∀a,bϵQ
∗ is commutativve.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=
2
(
2
ab
)∗c
=
4
abc
(a∗b)∗c=a∗(b∗c)=
2
a×
2
bc
=
4
abc
Since (a∗b)∗c=a∗(b∗c)∀a,b,cϵQ
∗ is an associative binary operation.
iv)On z
+
, define if a∗b=b∗a
a∗b=2
ab
b∗a=2
ba
=2
ab
Since a∗b=b∗a∀a,b,cϵz
+
∗ is commutative.
Check associative.
∗ is associative if $$
(a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(2
ab
)
∗
c=2
2
ab
c
a∗(b∗c)=a∗(2
ab
)=2
a2
bc
Since (a∗b)∗c
=a∗(b∗c)
∗ is not an associative binary operation.
v)On z
+
define a∗b=a
b
a∗b=a
b
,b∗a=b
a
⇒a∗b
=b∗a
∗ is not commutative.
Check associative
∗ is associative if $$
(a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(a
b
)
∗
c=(a
b
)
c
a∗(b∗c)=a∗(2
bc
)=2
a2
bc
eg:−Leta=2,b=3 and c=4
(a∗b)
∗
c=(2∗3)
∗
4=(2
3
)
∗
4=8∗4=8
4
a∗(b∗c)=2
∗
(3∗4)=2
∗
(3
4
)=2∗81=2
81
Since (a∗b)∗c
=a∗(b∗c)
∗ is not an associative binary operation.
vi)On R−{−1}, define a∗b=
b+1
a
Check commutative
∗ is commutative if a∗b=b∗a
a∗b=
b+1
a
b∗a=
a+1
b
Since a∗b
=b∗a
∗ is not commutatie.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(
b+1
a
)
∗
c=
c
b
a
+1
=
c(b+1)
a
a∗(b∗c)=a∗(
c+1
b
)=
c+1
b
a
=
b
a(c+1)
Since (a∗b)∗c
=a∗(b∗c)
∗ is not a associative binary operation
27 since 27 is a factor of 54
