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OK...table looks correct.

b   out of 1000 students   600 passed       so   600/1000 = 3/5 = 60 %

c  380 out of 450 males passed the exam     380/450  =   38/45 = 84% of males passed the exam.

d  220/550 =  22/55= 40% of females passed the exam

e  It DOES appear that if student is a male , student is more likely to pass the exam.

deleted......

Some info missing here...

Sum of a geometric series   Sn =  a1/(1-r)

in this geometric series   r = x     a1 = 4

if you want total to be greater than 8:

8< 4/(1-x)

8-8x < 4

4<8x

x> 1/2         Interval notation:     (1/2, 1)      we have to limit it to 1    r cannot equal 1 (the equation would have 0 in denominator) and it cannot be GREATER than one....because then the series becomes divergent and has no sum.

OK...we just get to make it up   

   Let's set a1 = 5

      and d=3

a1 = 5

a2 = 5 +3 = 8

a3 = 5+3+3  = 11

a4 = 5 +3+3+3 =14

a5 = 5 +3+3+3+3 = 17

an = 5 + (3)(n-1)

Edited...... Thanx CPhill for pointing out I answered with a GEOMETRIC series rather than an Arithmetic series !!!

35 out of 100 chose chocolate ice cream with a waffle cone  =  35/100 ==  7/20 ==  35%

A         a1* r^(n-1) =   -3 * -2^5 =  96

B  nth= 5* r^(n-1) = 5 * 2^(n-1)

c  if a3 =96  and r =2      then    a2 = a3/r = 48       a1 = a2/r =24      a7 =  a1*r(n-1) = 24*2^6 = 1536

A line with slope of -2 intersects the positive x-axis at A and the positive y-axis at B. A second line intersects the x-axis at C(8,0) and the y-axis at D. The lines intersect at E(4,4). What is the area of the shaded quadrilateral OBEC?

A line with slope of -2 intersects the positive x-axis at A and the positive y-axis at B. A second line intersects the x-axis at C(8,0) and the y-axis at D. The lines intersect at E(4,4). What is the area of the shaded quadrilateral OBEC?

Thanks for that Alan.  I had been following this factor theory ... Need to trace the fundamental mistake!

Regards and I re-assert that you do a great job !

For the inequality: y >= 1/2 (x - 3)^2 +1   identify the following:

            Vertex (0|1)

            Intercepts y=1

            The graph flip up.

            The graph of the line will be solid.

A frustum of a right circular cone is formed by cutting a small cone off of the top of a larger cone. If a particular frustum has an altitude of $24$ centimeters, the area of its lower base is $225\pi$ sq cm and the area of its upper base is $25\pi$ sq cm, what is the altitude of the small cone that was cut off?

For b:   \(\log_5w+\frac{\log_5u}{3}-\frac{\log_5v}{3}\rightarrow \log_5w+\frac{\log_5u/v}{3} \rightarrow \log_5w+\log_5((\frac{u}{v})^{1/3})\rightarrow \log_5(w(\frac{u}{v})^{1/3})\)

or you could express this as \(\frac{\log_5(w^3u/v)}{3}\)

For e you should have 2x, not just 2.

Prove as follows:

(Note: your expressions for a and b are incorrect).

\(2x - y = -3\\ 10x+7y = -3\)

\(\text{substitution}\\ y = 2x+3\\ 10x+7(2x+3) = -3\\ 24x+ 21=-3\\ 24x=-24\\ x=-1, y = 1\\ --\\ \text{elimination}\\ \text{subtract 5x the first equation from the second}\\ 12y=12\\ y=1,~x=-1\)

\(a_{n+2} = a_{n+1}+a_n,~n\geq 3\\ s = \sum \limits_{n=1}^{10} a_n = 55 a1 + 88 a2\\ a_7 = 5a_1+8a_2 = 6\\ a_2 = \dfrac{6-5a_1}{8}\\ s = 55a_1 + 88\dfrac{6-5a_1}{8} = 55a_1 + 66-55a_1 = 66\)

DRC and DCR are the same angles in the triangle (opposite side lengths are the same)

The inscribed angle D encompasses a 180 degree arc ....so angle D = 90 degrees

Triangle = 180 degrees total =  90 + DRC + DCR        DRC and DCR are equal  so they are each 45 degrees.

On August 3   distance will be    5km + .1 + .1 + .1 = 5.3km

   yes it is an arithmetic sequence because it increases at a constant rate ......each term differs from the previous term by a constant amount.

distance for nth term =   (5 + .1n)   km

If he starts at 5km on Aug 1st and THEN increases 0.1 km daily

    distance for nth day =  5 + .1(n-1)     on the n=16th day      distance = 5.15 km

i.  5% of 20,000 is 1,000, so initially the options are equal 

ii.  Take Option 1.  Not only is it more beneficial after 10 years, it's more beneficial after only 2 years.   After the first year, the salary is 21,000 so calculating the salary increase on 21,000 the second year the increase would be 5% of 21,000 which is 1,050.  That made the salary 22,050 and the next increase would be 5% of 22,050 which is 1,102.50.