Expert: Josh - 6/14/2004

Question
1) when administering a standard intelliegence test, we expect about 1/3 of the scores to be more than 12 units above 100 or more than 12 units below 100. Describe this situation by writing an absolute value inequality.

2) The manager of a large apartment complex has found that the profit is given by P=-x^250x-15,000, where x is the number of apartments rented. For what values of x does the complex produce a profit?

3)Solve the rational inequality x^2+x-2/x^2-2x-3<0

4) Find the equation for the line that passes through the point (-5,2) and parallel to the line taht passes throught the point (1,2) and (4,3)

5) Solve the inequality  8/3(x-4)<=2/9(3x+2)

Answer
Christysen Andall,

Q1. Use variable X to represent the outcome of an intelligence test. Let P(X>=x) be the probability of X greater than or equal to x.

The two thresholds are x_upper=100+12=112 and x_lower=100-12=88 respectively.

The situation is described by P(X>112)=1/3 and P(X<88)=1/3.
Meaning, probability of obtaining a score between 88 and 112 (inclusive) is given by P(88=<X=<112)=1/3.

Q2. Idea: Solve inequality for "x", with expression P being positive. eg., solve for a*x^2+b*x+c>0.
I think you copied the equation wrong. Cannot proceed.

Q3. Brackets! without them, the expression makes no sense.
(x^2+x-2)/(x^2-2x-3)<0
(x-1)(x+2)/[(x+1)(x-3)]<0
One way to understand this is to sketch two parabolas for the numerator and denominator (only the sign matters here) and consider their quotient. Then, determine which interval of x returns a negative value.

(i) Numerator parabola is U-shape and it intersects with the x-axis at x=1,x=-2.
i.e., (x^2+x-2) is positive from x=-infinity to -2 and from x=1 to infinity. (x^2+x-2) is negative between x=-2 to x=1, excluding the end points.

(ii) Denominator parabola is also U-shape, since leading coefficient is positive. It intersects with x-axis at x=-1,x=3. So, (x^2-2x-3) is positive from x=-infinity to -1 and from x=3 to infinity; but negative between x=-1 to 3.

We have several intervals to consider,
INTERVAL     NUMERATOR  / DENOMINATOR
(-inf,-2):      +       /      +
(-2,-1):        -       /      +
(-1,1):         -       /      -
(1,3):          +       /      -
(3,inf):        +       /      +

Therefore, the original expression is negative only inside the two intervals (-2,-1) and (1,3).

Q4. Fundamental: remember (y-y1)=m*(x-x1), where m=gradient (or slope of the straight line), (x1,y1) is one such point on the line.

Calculate gradient by m=(y3-y2)/(x3-x2), where (x2,y2)=(1,2), (x3,y3)=(4,3). Can do this, since the line passing through (x1,y1)=(-5,2) is parallel to the line passing through (x2,y2) and (x3,y3).

Q5. Cannot do this. Your syntax is ambiguous.
If you mean fraction 8/3 multiplied by (x-4), write (8/3)*(x-4). If you mean 8 divided by the product of 3*(x-4), write 8/[3*(x-4)].