The probability of occurrence for the events A, B and C is; 1/4.
<h3>What is the probability of occurrence of.the described events?</h3>
For the first event A in which case, there's no odd number on the first two rolls, the possible events are; EEE and EEO. Consequently, the required probability is;
Event A = 2/8 = 1/4.
For the event B in which case, there's an even number on both the first and last rolls; the possible events are; EEE and EOE. Consequently, the required probability is;
Event B = 2/8 = 1/4.
For the event C in which case, there's an odd number on each of the first two rolls; the possible events are; OOO and OOE. Consequently, the required probability is;
Event C = 2/8 = 1/4.
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Step-by-step explanation:
First you find out the area of the square.
Then you find the area of the circle.
After this you add the Two area together.
Answer:
Step-by-step explanation:
we know that
The Domain of the function are all the values of x that can take the function
in this problem the domain is
The Range of the function are all the values of y that can take the function
in this problem the range is
therefore
{y | y = –7, –1, 0, 9} baaaam
know my brain hurts i hope this helped because i really tried haha sorry if it doesnt
Answer:
.5299192646
Step-by-step explanation:
Just plug it in a calculator
Answer:
Tn = 2Tn-1 - Tn-2
Step-by-step explanation:
Before we can generate the recursive sequence, we need to find the nth term of the given sequence.
nth term of an AP is given as:
Tn = a+(n-1)d
If a17 = -40
T17 = a+(17-1)d = -40
a+16d = -40 ...(1)
If a28 = -73
T28 = a+(28-1)d = -73
a+27d = -73 ...(2)
Solving both equations simultaneously using elimination method.
Subtracting 1 from 2 we have:
27d - 16d = -73-(-40)
11d = -73+40
11d = -33
d = -3
Substituting d = -3 into 1
a+16(-3) = -40
a - 48 = -40
a = -40+48
a = 8
Given a = 8, d = -3, the nth term of the sequence will be
Tn = 8+(n-1) (-3)
Tn = 8+(-3n+3)
Tn = 8-3n+3
Tn = 11-3n
Given Tn = 11-3n and d = -3
Tn-1 = Tn - d... (3)
Tn-1 = 11-3n +3
Tn-1 = 14-3n
Tn-2 = Tn-2d...(4)
Tn-2 = 11-3n-2(-3)
Tn-2 = 11-3n+6
Tn-2 = 17-3n
From 3, d = Tn - Tn-1
From 4, d = (Tn - Tn-2)/2
Equating both common difference
(Tn - Tn-2)/2 = Tn - Tn-1
Tn - Tn-2 = 2(Tn - Tn-1)
Tn - Tn-2 = 2Tn-2Tn-1
2Tn-Tn = 2Tn-1 - Tn-2
Tn = 2Tn-1 - Tn-2
The recursive formula will be
Tn = 2Tn-1 - Tn-2
