<span>f(x) = x</span>² <span>+ 12x + 6 </span>→ y = x² + 12x + 6<span>
Let us convert the standard form into vertex form.
1) Complete the squares. Isolate x</span>² and x terms.
<span>y - 6 = x</span>² + 12x
<span>
2) Create the perfect square trinomial. Whatever number is added on one side must also be added on the other side.
y - 6 + 36 = x</span>² + 12x + 36<span>
y + 30 = (x + 6)</span>²
<span>y = (x + 6)</span>² - 30 ← Vertex form
<span>
To check:
y = (x + 6) (x + 6) - 30
y = x</span>² + 6x + 6x + 36 - 30
<span>y = x</span>² + 12x + 6<span>
The zero that could be added to the given function is 36, -36</span>
A=P(1+rt)
A/P=1+rt
A/P-1=rt
AP-P/Pr = t
A-1/r=t
Answer:
Null hypothesis: ∪ = No possible child abuse or neglect
Alternative hypothesis: Uₐ = Possible child abuse or neglect
Step-by-step explanation:
Null hypothesis: ∪ = No possible child abuse or neglect
Alternative hypothesis: Uₐ = Possible child abuse or neglect
A type I error occurs when you reject the null hypothesis when it is true. In this situation, a type I error occurs when you conclude on possible child neglect or abuse and place the child in protective custody
A type II error occurs when you accept the null hypothesis when it is false. In this instance, a type II error occurs when you conclude on no possible child abuse or neglect when there is and fail to remove the child from the home.
In this case, the type II error is the more serious error. Failure to remove the child when there is possible child abuse or neglect will lead to more detrimental effect. Although, the type I error is also serious, it is not so detrimental as the type II error.
600+30+7 if that is what you meant...
