Which statement is true about the extreme value of the given quadratic equation? y=-3x2+12x-33 A. The equation has a minimum val
1 answer:
Given quadratic equation : .
A quadratic equation represents a parabola.
We have leading coefficient is -3, a negative number.
So, it would be a downward parabola and it would have it's vertex point on the top on the parabola.
So, it would obviously have a maximum value.
Let us find the x-coordinate of the parabola first and then y-coordinate of the parabola.
x = -b/2a = -12/2(-3) = -12/-6 = 2.
Plugging x=in given equation, we get
y=-3(2)^2+12(2)-33 = -3(4) +24 -33
= -12 +24 -33 = -21.
Therefore, maximum value is -21.
Therefore, correct option is :
<h3>B. The equation has a maximum value with a y-coordinate of -21.</h3>You might be interested in
Answer:
Null hypothesis:
Alternative hypothesis:
Step-by-step explanation:
1) Data given and notation
represent the sample mean
represent the population standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 8000, the system of hypothesis are :
Null hypothesis:
Alternative hypothesis:
Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
(1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
4) P-value
Since is a one-side upper test the p value would given by:
5) Conclusion
If we compare the p value and the significance level assumed, for example we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than $8000.