1 Convert 12\frac{2}{3}12
3
2
to improper fraction. Use this rule: a \frac{b}{c}=\frac{ac+b}{c}a
c
b
=
c
ac+b
\frac{12\times 3+2}{3}\times 3\frac{1}{4}
3
12×3+2
×3
4
1
2 Simplify 12\times 312×3 to 3636
\frac{36+2}{3}\times 3\frac{1}{4}
3
36+2
×3
4
1
3 Simplify 36+236+2 to 3838
\frac{38}{3}\times 3\frac{1}{4}
3
38
×3
4
1
4 Convert 3\frac{1}{4}3
4
1
to improper fraction. Use this rule: a \frac{b}{c}=\frac{ac+b}{c}a
c
b
=
c
ac+b
\frac{38}{3}\times \frac{3\times 4+1}{4}
3
38
×
4
3×4+1
5 Simplify 3\times 43×4 to 1212
\frac{38}{3}\times \frac{12+1}{4}
3
38
×
4
12+1
6 Simplify 12+112+1 to 1313
\frac{38}{3}\times \frac{13}{4}
3
38
×
4
13
7 Use this rule: \frac{a}{b}\times \frac{c}{d}=\frac{ac}{bd}
b
a
×
d
c
=
bd
ac
\frac{38\times 13}{3\times 4}
3×4
38×13
8 Simplify 38\times 1338×13 to 494494
\frac{494}{3\times 4}
3×4
494
9 Simplify 3\times 43×4 to 1212
\frac{494}{12}
12
494
10 Simplify
\frac{247}{6}
6
247
11 Convert to mixed fraction
41\frac{1}{6}41
6
1
41 and 1/6
Answer:
12%
Step-by-step explanation:
From what is there it is nearly impossible to find an exact number so I would say C... hope this helps
Answer:
B ; C ; D
Step-by-step explanation:
Number of faces on a number cube = 6
Sample space = (1, 2, 3, 4, 5, 6)
P(1 then 0)
P(1) = 1/6 ; P(0) = 0
P(1 then 0) = 1/6 * 0 = 0
P(even number then odd number) :
P(even number) = 3/6 = 1/2
P(odd) = 3/6 = 1/2
P(even number then odd number) = 1/2 * 1/2 = 1/4
P(6 then 2) :
P(6) = 1/6 ; P(2) = 1/6 = 1/2
P(6 then 2) = 1/6 * 1/6 = 1/36
P(even number then 5) :
P(even) = 3/6 = 1/2
P(5) = 1/6
P(even number then 5) = 1/2 * 1/6 = 1/12
P(odd number then 2) :
P(odd) = 3/6 = 1/2
P(2) = 1/6
P(odd number then 2) = 1/2 * 1/6 = 1/12
Answer:
a) Q(-2,1) is false
b) Q(-5,2) is false
c)Q(3,8) is true
d)Q(9,10) is true
Step-by-step explanation:
Given data is 
is predicate that 
then 
. where 
are rational numbers.
a)
when 
Here 
that is satisfied. Then
this is wrong. since 
That is 
Thus 
is false.
b)
Assume 
.
That is 
Here 
that is this condition is satisfied.
Then
this is not true. since 
.
This is similar to the truth value of part (a).
Since in both satisfied and 
for both the points.
c)
if 
that is 
and 
Here 
this satisfies the condition .
Then 
This also satisfies the condition .
Hence 
exists and it is true.
d)
Assume 
Here 
satisfies the condition
Then 
satisfies the condition .
Thus, 
point exists and it is true. This satisfies the same values as in part (c)
