Is y=1-x to the second power linear or nonlinear
1 answer:
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Answer:
The values in the table, taking into account the quadratic equation, are:
- x -3 -2 -1 0 1 2 3 4
- y <u>16</u> 9 <u>4</u> 1 <u>0</u> 1 <u>4</u> 9
Step-by-step explanation:
To obtain the values of the table, you must use the quadratic equation given:
- <u>y = x^2 - 2x + 1</u>
Now, you must replace the x with the one that is above the value you want to find, in the first case, we're gonna replace the value x with -3:
- y = x^2 - 2x + 1
- y = (-3)^2 - 2*(-3) + 1
- y = 9 + 6 + 1
- <u>y = 16</u>
When x is -1
- y = x^2 - 2x + 1
- y = (-1)^2 - 2*(-1) + 1
- y = 1 + 2 + 1
- <u>y = 4</u>
When x is 1
- y = x^2 - 2x + 1
- y = (1)^2 - 2*(1) + 1
- y = 1 - 2 + 1
- <u>y = 0</u>
When x is 3:
- y = x^2 - 2x + 1
- y = (3)^2 - 2*(3) + 1
- y = 9 - 6 + 1
- <u>y = 4</u>
At last, the graph must be as the attached picture I give you, but <u><em>remember in y-axis you must use 1 cm as unit and in the x-axis you must use 2 cm as unit, in this form, the graph will not be so elongated as the picture I attach, It would be wider</em></u>.
Answer:
1568
Step-by-step explanation:
The computation of the minimum sample size is shown below:
Given that
Error E = 11
The population standard deviation =217
Confidence level 1-α = 95%
Now following formula should be used
n > (Zalpha ÷ 2 × sigma ÷ E)^2
As we know that
Zalpha ÷ 2= 1.96
Now the sample size is
n> (1.96 × 202 ÷ 10)^2
= 1568