Please help if you can!
1 answer:
9514 1404 393
Answer:
y = 1/2x^2 +x -15/2
Step-by-step explanation:
The vertex is the point where the graph reaches is extreme: (-1, -8). The y-intercept is (0, -7.5). The x-intercepts are (-5, 0) and (3, 0).
When the quadratic is written in the form ...
y = x^2 +bx +c
the value of b is the opposite of the sum of the x-intercepts. The value of c is their product. These facts would give ...
y = x^2 -(-5 +3)x +(-5)(3) = x^2 +2x -15
The constant in this equation is -15, which is the value y has when x=0. Our actual y-intercept is half that, so the graph we see is this equation scaled by a factor of 1/2:
y = 1/2x^2 +x -15/2
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<em>Alternate solution</em>
You can start with the vertex form.
y = a(x -h)^2 +k . . . . . . . vertical scale factor 'a' and vertex (h, k)
On the graph, the vertical scale factor is the vertical difference between the low point (vertex) and the point that is 1 unit right or left. Here, that is ...
-7.5 - (-8) = 0.5 = a
We've already said the vertex is (h, k) = (-1, -8), so the vertex form equation is ...
y = 0.5(x +1)^2 -8
We know the square of a binomial is (a+b)^2 = a^2 +2ab +b^2, so we have ...
y = 0.5(x^2 +2x +1) -8
y = 0.5x^2 +x +0.5 -8 . . . . . eliminate parentheses
y = 0.5x^2 +x -7.5
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Answer:
12.92%
Step-by-step explanation:
Mean of the scores= u = 500
Standard deviation = = 10.6
We have to find what proportion of students scored more than 512 marks.
The distribution of scores in a test generally follows the Normal distribution. So we can assume that the distribution of MCAT scores is normally distributed about the mean.
Since, the distribution is normal, we can use the concept of z scores to find the proportion of students who scored above 512.
The formula for z scores is:
So, z score for x = 512 will be:
Thus,
P(X > 512) is equivalent to P(z > 1.13)
So, the test scores of 512 is equivalent to a z score of 1.13. Using the z table we have to find the proportion of z scores being greater than 1.13, which comes out to be 0.1292
Since,
P(X > 512) = P(z > 1.13)
We can conclude that, the proportion of students taking the MCAT who had a score over 512 is 0.1292 or 12.92%