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Since 24% of 1200 students have a part-time job and on the honor roll,

there are \( 0.24 \cdot 1200= 288\) students that have a part-time job and on the honor roll.

there are \(0.62\cdot1200=744\) students that have a part-time job, including the ones on the honor roll.

Since the problem states that they selected a student with a part time job, it must be one of the 744 students.

Out of the 744, we need the student to be one of the 288 people, so the probabilty of this happening is:

\(288\div744\approx0.3871\)

Rounding this to the nearest percent, we have 39%.

your final answer will be b.

I hope this helped, 

Gavin.

Once again, I do not understand the method you are asking me to solve the problem with. 

Here is my method. 

Since the problem is asking for the percentage of students that pass edthe test, we first need to know how many students actually did passed the test. 

We need to find all the numbers above 60, from this list 76, 55, 73, 93, 29, 74, 57, 34, 54.

We have 76, 73, 93, and 74, the rest are all below 60.

Since there are a total of 9 students, and only 4 passed the test. 

The percentage of students that passed the test is 4/9, or \(0.4\overline4%\)%

I hope this helped,

Gavin.

I do not understand the method you want me to solve the problem with. 

This is mine:

We can first find the median of this list of data, then find the numbers that are most of.

First, we can arrange it from least to greatest:

\(29, 34 , 54, 55, 57, 73, 74, 76, 93\)

The median, or middle number is 57.

The numbers farthest away from 57 are:

\( |57 - 29| = 28 \)

\(|57-93|=-36\)

\(|57-34|=23\)

Therefore, the scores that are most unusual are:

29, 93, and 34.

I hope this helped,

Gavin

Here is my solutions.

4. To "light" up all the squares with a knight going from A1 to H8, the knight must travel 63 times.

Everytime a knight moves, it goes from black to white square and vise versa. 

To go from white to white, the knight must move an even amount of times. 

Therefore this is not possible. 

Wow! Well done!

1)  BEEPER.

6! / 3! = 120 Number of  distinct arrangements.

2) TENNESSEE.

9! / 4!.2!.2! =3,780 Number of distinct arrangements.

+124 

how do I find the value of x which satisfies both x - 3 >2 and 2x + 1 < 15

x - 3 >2

x-3+3>2+3

x>5

and

2x + 1 < 15

2x+1-1< 15-1

2x<14

x<7

x is bigger than 5 and less than 7 so

x is between 5 and 7

5 < x < 7

z = (x - x_av)/sd.  So 

Here we have: x = 105, x_av = 103, sd = 2.7

You can crunch the numbers!

 76, 55, 73, 93, 29, 74, 57, 34, 54, and 75 

Order the data from low to high

29  34  54  55  57  73  74  75  76  93  

The median  is :    [ 57 + 73] / 2   =  130/2   =  65

The mean is   [ 29 + 34 + 54 +  55 +  57 +  73  +  74 +  75 +  76 +  93]  / 10  = 62

The standard deviation  is ≈  20

We take the absolute value difference  of each score from the mean, square it  and add these squares together......then, for a sample,  we  divide this  by  the number of scores less one......the square root of the last calculation is the standard deviation

Let R  be the car's rate

Call the time for the car to travel  (in hours)  =  210/R

Call the total time for the bike to travel  (in hours)  =   210/R - 1/3

Now.....the distance bike travels   =  its rate * its time  =  120 *  [ (210/R) - 1/3 ] 

And the distance the  bike travels to meet the  car  = 120* [ (210/R - 1/3] / 2  =    60[ (210/R - 1/3 ] = the same distance the bike travels back to "A"

So.....the distance the bike travels back to "A" / ts rate  =  some time   (1)

And the distance the car travels after meeting the bike / its rate  =  the same time   (2)

So.......equating  (1)  and (2) in algebraic terms, we have

( 210 - 60 [ 210/R - 1/3] ) / R  =    60[ 210/R-  1/3 ] / 120

(210  - 60 [ 210/R - 1/3 ] ) / R  =   [ 210/ R - 1/3 ] / 2        cross-multiply

2(210 - 60[ 210/R - 1/3 ] =  R [ 210/R - 1/3 ]

420  - 120 [ 210/R - 1/3 ]  =  210  -  R /3 

420  - 25200/ R  +  40  =  210 - R/3

460 - 25200/R  = 210 - R/3 

R/3 + 250 - 25200/R  = 0        multiply though by R

(1/3)R^2 + 250R - 25200  = 0       multiply through by 3

R^2  + 750R  - 75600  = 0          factoring, we get

(R - 90) (R + 840)  = 0

Only the first factor set to 0  will provide a positive solution for R  =  90 kmh   =   the car's rate

For the events to be independent

P(A) * P(B)  =   P ( A and B)

( 1/4) * (2/3)  =  2 / 12  =  1/6

But  P ( A and B )  =   1/5

So....they are dependent events  and  "D"  is correct

Let  H  = the probability of eating a hot dog

Let S  = the probability of drinking a soda

So

P ( S  l  H )   =     P (S  and H)                .38    

                           ___________     =       ___    =   .62

                               P(H)                          .61

\(AB=BC \qquad\text{ABC is a right angled isosceles triangle}\\ 2*(AB)^2=(6\sqrt2)^2\\ 2*(AB)^2=72\\ (AB)^2=36\\ AB=BC=6\)

\(cos 60=\frac{1}{2}\)

\(\frac{x}{6}=\frac{1}{2}\\ x=3 \)

I suspect this means something like:

X is a matrix with, say, N rows and M columns.  Subscript i runs from 1 to N; subscript j runs from 1 to M.

\(\Sigma_iX_{ij}^2\) returns the sum of the squares of all elements from row 1 to row N in column j.

Hi Gavin,

I'll give this one a go :)

\(\frac{1}{\log_2\pi}+\frac{1}{log_5\pi}>2\)

\(LHS\\ =\frac{1}{\log_2\pi}+\frac{1}{log_5\pi}\\ =1\div \frac{\log_\pi\pi}{\log_\pi2}+1\div \frac{\log_\pi\pi}{\log_\pi5}\\ =\frac{\log_\pi2}{\log_\pi\pi}+ \frac{\log_\pi5}{\log_\pi\pi}\\ =\frac{\log_\pi2+\log_\pi5}{\log_\pi\pi}\\ =\frac{\log_\pi10}{1}\\ =\log_\pi10\\ \text{Now pi^2 is approx 9.87 which is less than 10 so}\\ =\log_\pi(\pi^2*\delta) \qquad \text{where delta is a real number slightly bigger than 1}\\ =\log_\pi(\pi^2) +\log_\pi \delta\\ =2+\text{a small positive number}\\ >2\)

QED

Well that is a part of a bigger question but it looks like it could a part of a matrix question.

It doesn't make sense on its own.

How many even perfect square factors does 2^4*7^9 have?

From the considerations below I  can see the answer is 10

         
\(7^2, \qquad 7^4, \qquad 7^6,\qquad 7^8 \qquad \text{none of these are even } \\~\\ \\ \text{all those below are even}\\ 2^2, \qquad 2^2*7^2, \qquad 2^2*7^4, \qquad 2^2*7^6,\qquad 2^2*7^8 \\ 2^4, \qquad 2^4*7^2, \qquad 2^4*7^4, \qquad 2^4*7^6,\qquad 2^4*7^8 \\ \)

Hi Mathcoder,

I see that you have viewed my answer, I am glad about that.    

But you have not given me any written feedback.                        

I did ask specific questions of you. It would be nice of you to respond. 

Your very much welcome!

I had to write my answer in multiple posts since LaTeX hates me right now. I might be wrong so make sure to double check my work. Good Luck!

If AB is 6, then 6 is the side of a 30-60-90 triangle or a half-cut equilateral triangle. This would mean that x is half of 6 or x is 3.

From this, you know that \(AB\Xi BC\) and the values are \(\frac{6\sqrt{2}}{\sqrt{2}} = 6.\)

The length of \(AB\) & \(BC\) are 6. This is because the hypotenuse of a right-isosceles triangle is \(\frac{h}{\sqrt{2}}.\)

HEEEEEEEEEEEEEELP!