Answer:
f(x) > 0 over the interval 
Step-by-step explanation:
If f(x) is a continuous function, and that all the critical points of behavior change are described by the given information, then we can say that the function crossed the x axis to reach a minimum value of -12 at the point x=-2.5, then as x increases it ascends to a maximum value of -3 for x = 0 (which is also its y-axis crossing) and therefore probably a local maximum.
Then the function was above the x axis (larger than zero) from 
, until it crossed the x axis (becoming then negative) at the point x = -4. So the function was positive (larger than zero) in such interval.
There is no such type of unique assertion regarding the positive or negative value of the function when one extends the interval from to -3, since between the values -4 and -3 the function adopts negative values.
-6z = 72
Divide both sides by -6
-6z = 72
/-6 /-6
z = -12
Answer:
D. If the P-value for a particular test statistic is 0.33, she expects results at least as extreme as the test statistic in exactly 33 of 100 samples if the null hypothesis is true.
D. Since this event is not unusual, she will not reject the null hypothesis.
Step-by-step explanation:
Hello!
You have the following hypothesis:
H₀: ρ = 0.4
H₁: ρ < 0.4
Calculated p-value: 0.33
Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
In this case, you have a 33% chance of getting a value as extreme as the statistic value if the null hypothesis is true. In other words, you would expect results as extreme as the calculated statistic in 33 about 100 samples if the null hypothesis is true.
You didn't exactly specify a level of significance for the test, so, I'll use the most common one to make a decision: α: 0.05
Remember:
If p-value ≤ α, then you reject the null hypothesis.
If p-value > α, then you do not reject the null hypothesis.
Since 0.33 > 0.05 then I'll support the null hypothesis.
I hope it helps!
