PLSSSS HELLPPPP MEEEEE I WILLL GIVEEE BRAINLIESSSTTT!!!!
2 answers:
Answer:
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Algebra I</u>
- Slope Formula:
Step-by-step explanation:
<u>Step 1: Define</u>
Point (6, 8)
Point (5, 5)
<u>Step 2: Find slope </u><em><u>m</u></em>
Simply plug in the 2 coordinates into the slope formula to find slope<em> m</em>
- Substitute in points [SF]:
- [Fraction] Subtract:
- [Fraction] Divide:
You might be interested in
We define the probability of a particular event occurring as:
What are the total number of possible outcomes for the rolling of two dice? The rolls - though performed at the same time - are <em>independent</em>, which means one roll has no effect on the other. There are six possible outcomes for the first die, and for <em>each </em>of those, there are six possible outcomes for the second, for a total of 6 x 6 = 36 possible rolls.
Now that we've found the number of possible outcomes, we need to find the number of <em>desired</em> outcomes. What are our desired outcomes in this problem? They are asking for all outcomes where there is <em>at least one 5 rolled</em>. It turns out, there are only 3:
(1) D1 - 5, D2 - Anything else, (2), D1 - Anything else, D2 - 5, and (3) D1 - 5, D2 - 5
So, we have
probability of rolling at least one 5.
What are the total number of possible outcomes for the rolling of two dice? The rolls - though performed at the same time - are <em>independent</em>, which means one roll has no effect on the other. There are six possible outcomes for the first die, and for <em>each </em>of those, there are six possible outcomes for the second, for a total of 6 x 6 = 36 possible rolls.
Now that we've found the number of possible outcomes, we need to find the number of <em>desired</em> outcomes. What are our desired outcomes in this problem? They are asking for all outcomes where there is <em>at least one 5 rolled</em>. It turns out, there are only 3:
(1) D1 - 5, D2 - Anything else, (2), D1 - Anything else, D2 - 5, and (3) D1 - 5, D2 - 5
So, we have
Answer:
0
Step-by-step explanation:
given that we roll a fair die repeatedly until we see the number four appear and then we stop.
the number 4 can appear either in I throw, or II throw or .... indefinitely
So X = the no of throws can be from 1 to infinity
This is a discrete distribution countable.
Sample space= {1,2,.....}
b) Prob ( 4 never appears) = Prob (any other number appears in all throws)
=
where n is the number of throws
As n tends to infinity, this becomes 0 because 5/6 is less than 1.
Hence this probability is approximately 0
Or definitely 4 will appear atleast once.