Equation of a parabola with vertex at (2, -1) is
y = a(x - 2)^2 - 1
Using the given point: -3 = a(4 - 2)^2 - 1
-2 = a(2)^2
4a = -2
a = -1/2
Therefore, required equation is
y = -1/2(x - 2)^2 - 1
y = -1/2(x^2 - 4x + 4) - 1
y = -1/2x^2 + 2x - 2 - 1
y = -1/2x^2 + 2x - 3
Had to look for the options for this question and here is my answer.
Based on the given scenario above regarding Heather school's scheduling, I can say that the question that would be the most appropriate for her to ask is "How do you feel about your child having to take summer school in order to graduate based on this traditional schedule?" Hope this helps.
P + s = 200......p = 200 - s
20p + 15s = 3400
20(200 - s) + 15s = 3400
4000 - 20s + 15s = 3400
-20s + 15s = 3400 - 4000
-5s = - 600
s = -600/-5
s = 120 <=== there were 120 standard tickets sold
p + s = 200
p + 120 = 200
p = 200 - 120
p = 80 <=== there were 80 premium tickets sold
Answer:
The 95% confidence interval for the mean is between 0.985g/cm² and 1.047 g/cm².
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so 
Now, find M as such
In which 
is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the mean subtracted by M. So it is 1.016 - 0.0310 = 0.985 g/cm²
The upper end of the interval is the mean added to M. So it is 1.016 + 0.0310 = 1.047 g/cm²
The 95% confidence interval for the mean is between 0.985g/cm² and 1.047 g/cm².
A) First, we need to calculate x. We can do this by doing: 180-(70+50), which is nothing but 60.
Since the largest angle is 60, it means that the side opposite of it is the largest side. Which in this case, side AB.
So the largest side is AB.
Now, let's take a look at the smallest angle which is <B, because AC is the opposite of this angle, it must also be the smallest side.
We're left with angle <50, which is "medium" sized? Obviously, this can just be put in the middle (Side BC)
So, for response A, you should get AC, BC, AB.
Now for problem B, it's the same steps we did above. If you replicate what we did you should get PQ, PR,RQ.
Please let me know how I did.
Happy studying!
