Answer: B and F I believe
Step-by-step explanation:
Answer:

Step-by-step explanation:
<u>Substitute g(-3) into the function:</u>

<u>Multiply:</u>
<u />
<u>Add:</u>
<u />
The first set:
3x + 2y = 2 ---1)
5x + 4y = 6 ---2)
From 1), multiply all by 2, 6x + 4y = 4 ---3)
3) - 2),
6x + 4y - (5x + 4y) = 6 - 4
6x + 4y - 5x - 4y = 2
x = 2
Sub in x = 2 into 1),
3(2) + 2y = 2
2y = -4
y = -2
(2 , -2)
The second set:
3x + 2y = 2 ---1)
11x + 8y = 10 ---2)
From 1), multiply all by 4, 12x + 8y = 8 ---3)
3) - 2),
12x + 8y - (11x + 8y) = 8 - 10
12x + 8y - 11x - 8y = -2
x = -2
From this x value alone, we can tell that these two linear systems do NOT have the same solution as they meet at different coordinates.
Hope this helped! Ask me if there's any working from here that you don't understand! :)
Answer:
6/25
Step-by-step explanation:
its a probability out of all the odd numbers such as 1,3,5,7,9,11,13,15,17,19,21,23,25
Answer:
a)0.08 , b)0.4 , C) i)0.84 , ii)0.56
Step-by-step explanation:
Given data
P(A) = professor arrives on time
P(A) = 0.8
P(B) = Student aarive on time
P(B) = 0.6
According to the question A & B are Independent
P(A∩B) = P(A) . P(B)
Therefore
&
is also independent
= 1-0.8 = 0.2
= 1-0.6 = 0.4
part a)
Probability of both student and the professor are late
P(A'∩B') = P(A') . P(B') (only for independent cases)
= 0.2 x 0.4
= 0.08
Part b)
The probability that the student is late given that the professor is on time
=
=
= 0.4
Part c)
Assume the events are not independent
Given Data
P
= 0.4
=
= 0.4

= 0.4 x P
= 0.4 x 0.4 = 0.16
= 0.16
i)
The probability that at least one of them is on time
= 1-
= 1 - 0.16 = 0.84
ii)The probability that they are both on time
P
= 1 -
= 1 - ![[P({A}')+P({B}') - P({A}'\cap {B}')]](https://tex.z-dn.net/?f=%5BP%28%7BA%7D%27%29%2BP%28%7BB%7D%27%29%20-%20P%28%7BA%7D%27%5Ccap%20%7BB%7D%27%29%5D)
= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56