Answer:
1/3 or 0.333333
Step-by-step explanation:
The question talks about a die that was tossed once and we are now asked to find the probability that it is a figure greater than 4
---A die has 6 faces with the numbers 1,2,3,4,5 and 6 one on each of
the six faces.Considering an unbaised die the probability of any of the faces to appear in a single roll is 1/6.
So, P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=1/6
Now that numbers greater than 4 are 5 and 6
So it can be P(5) or P(6)= P(5)+P(6)=2/6 = 1/3 or 0.33333
Answer 1/3 or 0.33333
Good evening
Answer:
<h2>-3y² − 4y - 6</h2><h2 />
Step-by-step explanation:
______________________
y² – 5y + 1 − P = 4y² – y + 7
then
P = y² – 5y + 1 − (4y² – y + 7)
then
P = y² – 5y + 1 − 4y² + y − 7
then
P = (y² – 4y²) − (5y − y) + (1 − 7)
then
P = -3y² − 4y - 6
_______________________
:)
14,7
Each number is twice smaller the the previous one.
