Raphael graphed the functions g(x) = x + 2 and f(x) = x – 1. How many units below the y-intercept of g(x) is the y-intercept of
f(x)?
2 answers:
Answer:
The distance between y-intercepts of f(x) and g(x) = 3 units
Step-by-step explanation:
For better understanding of the solution, see the plotted graph which is attached below :
f(x) = x - 1
g(x) = x + 2
Now, to find the distance between the y - intercepts of f(x) and g(x), First find the y-intercepts of both the given functions
y-intercepts of f(x) = x - 1
At y-intercept the x-coordinate = 0
⇒ y = 0 - 1
⇒ y = -1
y-intercepts of g(x) = x + 2
At y-intercept the x-coordinate = 0
⇒ y = 0 + 2
⇒ y = 2
Now, distance between the two intercepts = 2 - (-1)
= 2 + 1
= 3
Hence, the distance between y-intercepts of f(x) and g(x) = 3 units
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Step-by-step explanation:
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Answer and explanation:
To find : List the sample space and tell whether you think the events are equally likely ?
Solution :
a) Toss 2 coins; record the order of heads and tails.
Let H is getting head and t is getting tail.
When two coins are tossed the sample space is {HH,HT,TH,TT}.
Total number of outcome = 4
As the outcome HT is different from TH. Each outcome is unique.
Events are equally likely since their probabilities are same.
b) A family has 3 children; record the number of boys.
Let B denote boy and G denote girl.
If there are 3 children then the sample space is
{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}
The possible number of boys are 0,1,2 and 3.
Number of boys Favorable outcome Probability
0 GGG
1 GGB,GBG,BGG
2 GBB,BGB,BBG
3 BBB
Since the probabilities are not equal the events are not equally likely.
c) Flip a coin until you get a head or 3 consecutive tails; record each flip.
Getting a head in a trial is dependent on the previous toss.
Similarly getting 3 consecutive tails also dependent on previous toss.
Hence, the probabilities cannot be equal and events cannot be equally likely.
d) Roll two dice; record the larger number
The sample space of rolling two dice is
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Now we form a table that the number of time each number occurs as maximum number then we find probability,
Highest number Number of times Probability
1 1
2 3
3 5
4 7
5 9
6 11
Since the probabilities are not the same the events are not equally likely.