First of all, you need to come to an understanding of what you mean by "compare that score to the population." Often, that will mean determining the percentile rank of the score.
To determine the percentile rank of a raw score, you first nomalize it by determining the number of standard deviations it lies from the mean. That is, you subtract the population mean and divide that difference by the population standard deviation. Now, you have what is referred to as a "z-score".
Using a table of standard normal probability functions (or an equivalent calculator or app), you look up the cumulative distribution value corresponding to the z-score you have. This number between 0 and 1 (0% and 100%) will be the percentile rank of the score, the fraction of the population that has raw scores below the raw score you started with.
Answer:
300
Step-by-step explanation:
6 × 10¹ ÷ 2 × 10-¹
= 300
no of aluminium is 300 times more than that of gold
If two angles are complementary, the sum of their measures is 90 degrees. Call one of the angles x. This means that its complement is 90 - x degrees. But the difference in their measures is 12 degrees.
Answer:
a. A(x) = (1/2)x(9 -x^2)
b. x > 0 . . . or . . . 0 < x < 3 (see below)
c. A(2) = 5
d. x = √3; A(√3) = 3√3
Step-by-step explanation:
a. The area is computed in the usual way, as half the product of the base and height of the triangle. Here, the base is x, and the height is y, so the area is ...
A(x) = (1/2)(x)(y)
A(x) = (1/2)(x)(9-x^2)
__
b. The problem statement defines two of the triangle vertices only for x > 0. However, we note that for x > 3, the y-coordinate of one of the vertices is negative. Straightforward application of the area formula in Part A will result in negative areas for x > 3, so a reasonable domain might be (0, 3).
On the other hand, the geometrical concept of a line segment and of a triangle does not admit negative line lengths. Hence the area for a triangle with its vertex below the x-axis (green in the figure) will also be considered to be positive. In that event, the domain of A(x) = (1/2)(x)|9 -x^2| will be (0, ∞).
__
c. A(2) = (1/2)(2)(9 -2^2) = 5
The area is 5 when x=2.
__
d. On the interval (0, 3), the value of x that maximizes area is x=√3. If we consider the domain to be all positive real numbers, then there is no maximum area (blue dashed curve on the graph).