Answer:
400 is the answer
Step-by-step explanation:
Answer:
The desired z-score is 2.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
n(12,2.5)
This means that 
z score when x =17
This is Z when X = 17. So



The desired z-score is 2.
We can check this two ways: distributing the -0.5, and plugging in a number for x.
First, we can check this by distributing the -0.5. Our equation becomes:
-1.5x -2.5 = -1.5x + 2.5
Clearly, this is not true. We can show this one more time by plugging in a number for x. I will use 2.

Now, we evaluate both sides and see if they are equal.
-3 - 2.5 = -3 + 2.5
-5.5 = 0.5
Clearly, this is not true. These expressions are not equivalent.
Triangle ABC is similar to triangle CEF.
<u>Explanation:</u>
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Diagram is inserted for the reference.
ABCD is a rectangle.
ABC is a right angled triangle because all the angles of the rectangle are 90◦ - (a)
CEF is a right angled triangle because FE is perpendicular to DC – (b)
In triangles ABC and CEF,
1. Angle ABC = Angle CEF = 90◦ (Both are right angles from a and b)
2. Angle BCA = Angle EFC (Alternate angles on parallel lines are equal on intersection)
Hence using Similarity property of AA (Angle, Angle), Triangle ABC and CEF are similar.
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From the double-angle identity,

we can rewritte our given equation as:

By factoring 2cosx on the left hand side, we have

This equation has 2 solutions when

From equation (A), we obtain

and from equation (B), we have

On the other hand, we can find one more solution from the original equation by substituting x=0, that is,

then, x=0 is another solution. In summary, we have obtained the following solutions:

However, the intersection of the last set is empty. So the unique solution is x=0 as we can corroborate on the following picture:
Therefore, the solution set is: {0}