Question 5!!!!!!!!!!!!!!!!
1 answer:
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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
Step-by-step explanation:
Given that,
a)
X ~ Bernoulli and Y ~ Bernoulli
X + Y = Z
The possible value for Z are Z = 0 when X = 0 and Y = 0
and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0
If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,
Therefore Z is a Bernoulli random variable
b)
If X and Y can not be both equal to 1
then,
or
and
c)
If both X = 1 and Y = 1 then Z = 2
The possible values of the random variable Z are 0, 1 and 2.
since a Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution