We need to define our outcomes and events.
Finding the probability<span> of each event occurring
separately, and then multiplying the probabilities is the step to <span>finding
the probability</span> of two
independent events that occur in
sequence.
</span>
<span>
To solve this problem, we take note of this:</span>
The roll of the two dice are denoted by the pair
(I, j) ∈ S={ (1, 1),(1, 2),..., (6,6) }
Each pair is an outcome. There are 36 pairs and each has
probability 1/36. The event “doubles” is { (1, 1),(2, 2)(6, 6) } has
probability p= 6/36 = 1/6. If we define ”doubles” as a successful roll, the
number of rolls N until we observe doubles is a geometric (p) random variable
and has expected value E[N] = 1/p = 6.
Answer:
x = 3??
Step-by-step explanation:
1) 8 + 4 = -5 + 7
12 = 2
FALSE
2) y = -11x + 4
(0, -7): -7 = -11(0) + 4 ⇒ -7 = 0 + 4 ⇒ -7 = 4 False
(-1, -7): -7 = -11(-1) + 4 ⇒ -7 = 11 + 4 ⇒ -7 = 15 False
(1, -7): -7 = -11(1) + 4 ⇒ -7 = -11 + 4 ⇒ -7 = -7 True
(2, 26): 26 = -11(2) + 4 ⇒ 26 = -22 + 4 ⇒ 26 = -18 False
Answer: C
3) Input Output
0 0
<u> 1 </u> 3
2 <u> 6 </u>
3 9
<u> 4 </u> <u> 12 </u>
5 15
6 <u> 18 </u>
Rule: input is being added by 1, output is 3 times x
4) c = 65h
5) 2x = -6
x = -3
6) 8j - 5 + j = 67
9j - 5 = 67 <em>added like terms (8j + j)</em>
<u> +5</u> <u>+5 </u>
9j = 72
j = 8
7) y = mx + b
<u> -b</u> <u> -b </u>
y - b = mx
I believe it’s 3 to the 34 power
A. 11/30 x 50= 18/36=1/2
18/36 x 1/10= 5/100
B. 1,210 x 1.1= 1,331
I am not so sure about A, but I hope this helps a little.
