F (x) = x^2 - 2x + 1
f (x) = (-2)^2 - 2(-2) + 1
= 4 + 4 + 1
= 9
f (x) = ( 0 )^2 - 2 (0) + 1
= 0 - 0 + 1
= 1
0 = x^2 - 2x + 1
x^2 = 2x - 1 = 1
f (2) = 2^2 - 2(2) + 1
= 4 - 4 + 1
= 1
f (3) = 3^2 - 2(3) + 1
= 9 - 6 + 1
= 3 + 1
= 4
Answer:
is this in graph or equation? ???
Answer:
0.1225
Step-by-step explanation:
Given
Number of Machines = 20
Defective Machines = 7
Required
Probability that two selected (with replacement) are defective.
The first step is to define an event that a machine will be defective.
Let M represent the selected machine sis defective.
P(M) = 7/20
Provided that the two selected machines are replaced;
The probability is calculated as thus
P(Both) = P(First Defect) * P(Second Defect)
From tge question, we understand that each selection is replaced before another selection is made.
This means that the probability of first selection and the probability of second selection are independent.
And as such;
P(First Defect) = P (Second Defect) = P(M) = 7/20
So;
P(Both) = P(First Defect) * P(Second Defect)
PBoth) = 7/20 * 7/20
P(Both) = 49/400
P(Both) = 0.1225
Hence, the probability that both choices will be defective machines is 0.1225
Linear because x moves up by 3 and y moves up by 7
Answer:
6 "surprise" flavors jellybeans
Step-by-step explanation:
From the information given:
The bag contained a mixture of regular flavored jellybeans and "surprise" jellybeans.
There are 40 jellybeans in the bag.
If 15% of the jellybean were "surprise" flavors.
Then, the number of expected "surprise" jellybeans will be:
= 15% of 40 jellybeans
= (15/100) × 40
= 6 "surprise" flavors jellybeans
If 6 surprise jellybean is contained in the bag;
Thus, the number of regular flavored jellybean will be
= 40 - 6
= 34 regular flavored jellybean
