<h2>
Answer:</h2>
Ques 1)
Ques 2)
<h2>
Step-by-step explanation:</h2>
Ques 1)
We know that if a graph is stretched by a factor of a then the transformation if given by:
f(x) → a f(x)
Also, we know that the translation of a function k units to the right or to the left is given by:
f(x) → f(x+k)
where if k>0 then the shift is k units to the left
and if k<0 then the shift is k units to the right.
Here the graph of f(x) is transformed into the graph of g(x) by a vertical stretch of 4 units and a translation of 4 units right.
This means that the function g(x) is given by:
Ques 2)
We know that the transformation of the type:
f(x) → f(x)+k
is a shift or translation of the function k units up or down depending on k.
If k>0 then the shift is k units up.
and if k<0 then the shift is k units down.
Here, The graph of the function f(x)=|3x| is translated 4 units up.
This means that the transformed function g(x) is given by:
Answer:
Difference: ¨c-d¨ variable: ¨a¨ coefficient: ¨9¨
Step-by-step explanation:
Just trust me bro
Answer:
Yes, but it is not linear. It is exponential : 
Step-by-step explanation:
On the first day we have 5 bacteria.
On the second day we will have 5*2 = 10 bacteria.
On the third day we will have 10*2 = 5*2*2 = 5*2^2 = 20 bacteria.
On the fourth day we will have 20*2 = 5*2*2*2 = 5*2^3 = 40 bacteria.
We can see that
,
where x is a number of days and f(x) gives us the number of bacteria.
A(bx − c) ≥ bc, implies (bx − c) ≥ bc /a and then bx ≥ bc/a + c, x<span>≥ c/a +c/b
so the solution is </span><span>3. [c/a + c/b, infinity)</span>
Answer:
P(A) = 0.2
P(B) = 0.25
P(A&B) = 0.05
P(A|B) = 0.2
P(A|B) = P(A) = 0.2
Step-by-step explanation:
P(A) is the probability that the selected student plays soccer.
Then:
P(B) is the probability that the selected student plays basketball.
Then:
P(A and B) is the probability that the selected student plays soccer and basketball:
P(A|B) is the probability that the student plays soccer given that he plays basketball. In this case, as it is given that he plays basketball only 10 out of 50 plays soccer:
P(A | B) is equal to P(A), because the proportion of students that play soccer is equal between the total group of students and within the group that plays basketball. We could assume that the probability of a student playing soccer is independent of the event that he plays basketball.
