Answer:
See below
Step-by-step explanation:
<em>On the graphs we see transformations of exponential functions</em>
<h3>Graphic 1 = Horizontal shift </h3>
- f(x) = 2ˣ is the parent function
- g(x) = 2ˣ⁺³ indicates shift to the left by 3 units
- h(x) = 2ˣ⁻¹ indicates the shift to the right by 1 unit
<h3>Graphic 2 =Vertical shift</h3>
- p(x) = (1/3)ˣ is the parent function
- r(x) = (1/3)ˣ⁺³ indicates shift up by 3 units
- q(x) = (1/3)ˣ⁻² indicates the shift down by 2 units
Answer:
a)120
b)6.67%
Step-by-step explanation:
Given:
No. of digits given= 6
Digits given= 1,2,3,5,8,9
Number to be formed should be 3-digits, as we have to choose 3 digits from given 6-digits so the no. of combinations will be
6P3= 6!/3!
= 6*5*4*3*2*1/3*2*1
=6*5*4
=120
Now finding the probability that both the first digit and the last digit of the three-digit number are even numbers:
As the first and last digits can only be even
then the form of number can be
a)2n8 or
b)8n2
where n can be 1,3,5 or 9
4*2=8
so there can be 8 three-digit numbers with both the first digit and the last digit even numbers
And probability = 8/120
= 0.0667
=6.67%
The probability that both the first digit and the last digit of the three-digit number are even numbers is 6.67% !
I think the answer is 7.6m
