3(2x+5)=3
3 • 2x] + [3 • 5] = 3
6x + 15 = 3
6x = –12
x = –2
2x – 2(3x – 2) = 2(x –2) + 20
2x – 2(3x – 2) = 2(x –2) + 20
2x – 6x + 4 = 2x – 4 + 20
– 4x + 4 = 2x + 16
–4x + 4 – 4 –2x = 2x + 16 – 4 -2x
–6x = 12
x = –2
The answer is B part
As the requirement for the student is either a senior or one who studies maths or both.
So, just add the values within the two venn circles divide it by total no of values.
=> 240+40+110/240+40+110+840
=> 360/1200
Y = -3x + b
4 = -3(-2) + b
4 = 6 + b
-2 = b
Answer:
H0 : p = 0.75 against H1: p > 0.75 One tailed test.
Step-by-step explanation:
We state our null and alternative hypotheses as
H0 : p = 0.75 against H1: p > 0.75 One tailed test.
In this case H0 is not defined as p≤ 0.75 because the acceptance and rejection regions cannot be set up. Therefore we take the exact value of H0 : p= 0.75.
The claim is that the probability of the workers getting their job through the internet is greater than 75% or 0.75.
As H0 is supposed to be less than we choose H1 to be greater than equality.