Step-by-step explanation:
you don't show us the choices.
anyway, the real graph must be similar to the one you are showing, as it goes to - and + infinity for x = 2.
because the denominator "x-2" will be 0 for x = 2.
but the horizontal limits are both y = +3 (and not 0).
because (3x-2)/(x-2) goes more and more to 3/1 the larger (or smaller in the - direction) x gets.
<h2>
Hello!</h2>
The answers are:
A.
and 
D.
and 
<h2>
Why?</h2>
To find which of the following pairs of numbers contains like fractions, we must remember that like fractions are the fractions that share the same denominator.
We are given two fractions that are like fractions. Those fractions are:
Option A.
and
We have that:
So, we have that the pairs of numbers
and
Share the same denominator, which is equal to 6, so, the pairs of numbers contains like fractions.
Option D.
and
We have that:
So, we have that the pair of numbers
and
Share the same denominator, which is equal to 7, so, the pairs of numbers constains like fractions.
Also, we have that the other given options are not like fractions since both pairs of numbers do not share the same denominator.
The other options are:
and
We can see that both pairs of numbers do not share the same denominator so, they do not contain like fractions.
Hence, the answers are:
A.
and
D.
and
Have a nice day!
Answer:
9. x>-4 or x≥1
10. a<2 or a≥-5
11. v≤7 or v≥-4
12. k≥5 or k<8
13. n>6.8 and n≤9
Step-by-step explanation:
9. -2x-7>1
-2x>8
x>-4
x-2≥-1
x≥1
10.a/-2 <-1
a<2
-4a+3≥23
-4a≥20
a≥-5
11. 6v+38≤-4
6v≤-42
v≤7
2(v+3)≥-2
2v+6≥-2
2v≥-8
v≥-4
12. 4(1-k)≥-16
4-4k≥-16
-4k≥-20
k≥5
7-6k<-41
-6k<-48
k<8
13. 10n-9>-59
10n>-68
n>6.8
n-6≤3
n≤9
Are you looking for y and if so y=80
Correction:
Because F is not present in the statement, instead of working onP(E)P(F) = P(E∩F), I worked on
P(E∩E') = P(E)P(E').
Answer:
The case is not always true.
Step-by-step explanation:
Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.
And for any two mutually exclusive events, E and E',
P(E∩E') = 0
Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then
P(E)P(E') cannot be equal to zero.
So
P(E)P(E') ≠ 0
This makes P(E∩E') different from P(E)P(E')
Therefore,
P(E∩E') ≠ P(E)P(E') in this case.
