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## #1 Re: Help Me ! » Finding vertexes » 2013-12-04 13:25:42

I couldn't figure out how to find the vertexes so I tried a different approach. I think I got it.

Step 1
Row Operation    X           Y          U    V    W    M
Type A         -50          -300           1    0    0    -1400
Type B           3           4           0    1    0    42
14000      90000      0    0    1    0

Pivot on -50

Row Operation    X    Y          U    V    W    M
Type A            1    6        -1/50    0    0    28
Type B           0    -14         3/50    1    0    -42
0    6000      280    0    1    0

Pivot on -14

Row Operation    X    Y    U           V          W    M
Type A            1    0    1/175    3/7           0    10
Type B            0    1    -3/700    -1/14       0    3
0    0    2410/7    3000/7     1    -410000

10x+3y
To minimize the cost of the trip, the airline should fly 10 type A planes and 3 type B planes.

## #2 Help Me ! » Finding vertexes » 2013-12-04 10:11:49

mom
Replies: 3

I have the following problem. I do not know why I am stuck but I am can someone please help?

1.    An airline company wants to fly 1400 members of a ski club to Colorado.  The airline owns two types of planes. Type A can carry 50 passengers, requires 3 flight attendants, and costs \$14,000 for the trip.  Type B can carry 300 passengers, requires 4 flight attendants, and costs \$90,000 for the trip.  If the airline must use at least as many type A planes as type B and has available only 42 flight attendants, how many of each type should be used to minimize the cost for the trip.

For the constraints I got:

50x + 300y > 1400
3x + 4y < 42
x > y
x > 0, y > 0

I got standard form to be this:

y > -1/6x + 14/3
y < -3/4x + 21/2
y < x
x > 0, y > 0

With only two standard real constraints, how do I get three vertexes? Also I have big fractions which creates a challenge. Am I missing something?

## #3 Re: Help Me ! » balls in an urn » 2013-06-11 10:55:47

sorry was working on a group project. Yes it was 15/28

Thanks again

## #4 Re: Help Me ! » balls in an urn » 2013-06-11 08:22:04

I did thanks again

## #5 Re: Help Me ! » balls in an urn » 2013-06-11 08:19:15

that gives a better image than how they are teaching me. Thank you. I can see it now. They have us do a tree with fractions and I get confused easily.

## #6 Help Me ! » card probability » 2013-06-11 08:07:10

mom
Replies: 2

what is the probability of getting either a black card or an ace in one draw from an ordinary deck of 52 cards?

I got 8/13
26/52 black cards + 4/52 aces=8/13

the answer that was marked correct is 7/13

What did I do wrong?

## #7 Re: Help Me ! » balls in an urn » 2013-06-11 08:01:42

I have 3/7 and 2/7 for the second draw. The first was 3/8
I know that I am doing something wrong

## #8 Re: Help Me ! » balls in an urn » 2013-06-11 07:52:31

I drew the tree. What step did I miss in the third problem?

## #9 Help Me ! » balls in an urn » 2013-06-11 07:33:02

mom
Replies: 11

An urn contains 3 yellow marbles and 5 green marbles. One marble is removed, its color noted and not replaced. A second marble is removed and its color noted.

What is the probability that both marbles are yellow
Pr(YY)=3/28
I did 3/8*2/7=3/28
What is the probability that they are both green?
Pr(GG)=5/14
I did 5/8*4/7=5/14
what is the probability that exactly one marble is yellow?
there were two chances of this
I did 3/8*5/7=15/56 and 5/8*3/7=15/56
Do I divide by two for the probability that exactly one marble is yellow?

I am not sure if I have done all of the steps for any of the questions. Can someone look over my work and give me advice?

## #10 Re: Help Me ! » Probability again » 2013-06-11 05:59:58

wow!! finally got my brain to work properly. I got it figured out. thanks

## #11 Re: Help Me ! » Probability again » 2013-06-11 02:34:29

Bobby,

The diagrams help tremendously but I am still drawing a blank for some reason. I did figure the following from using the visual.

250 boys, 120 play baseball, 130 play soccer, 60 play both.

Did not play either sport: 60+60+70=190  250-190=60  60/250=.24 probability

play exactly one sport: 60+70=130/250=.52 probability

play soccer but not baseball: 70/250=.28 probability

play soccer, given that he played baseball: 60/120=.50

played baseball, given that he did not play soccer: 60/120=.50

did not play baseball, given that he did not play soccer: the answer is .46 but for some reason, I can not visualize what to do in order to come to that conclusion. Could you please help?

no??

## #13 Re: Help Me ! » probability » 2013-06-10 15:10:15

thanks. No worries here. I am moving on till I can get help with that problem.

thanks again

## #14 Re: Help Me ! » probability » 2013-06-10 14:53:49

I do on that one and I did appreciate the joke haha just stressed out over this math. I posted another question as I am stuck yet again. UGH!!!

## #15 Help Me ! » Probability again » 2013-06-10 14:39:43

mom
Replies: 11

I am stuck again. I have:

Out of 250 third grade boys, 130 played baseball, 130 played soccer, and 60 played both. Find the probability that a boy chosen at random did not play either sport.

Consider event E, which is "a chosen boy satisfies the requirements". The probability of E is determined by the following formula.

Pr(E)=[number of outcomes in E]/N

Where the total number of outcomes N is equal to the number of boys. So the numerator is the number of boys who satisfy the requirements and the denominator is the total number of boys.

Let A be the set of boys who play baseball, N(A) - number of boys who play baseball and B be the set of boys who play soccer, N(B) - number of boys who play soccer.

So, from the problem statement N(A)=130, N(B)=130,N(A and(upside down U) B)=60.

to find the probability that a boy chosen at random did not play either sport we should divide the number of such boys by the total number of boys. What is the number of boys who don't play either sport?

The answer is 50 but I don't know how to get to that answer. This is where I am lost so far. Can you please assist?

## #16 Re: Help Me ! » probability » 2013-06-10 12:44:33

I don't have to understand everything but when I am receiving a grade on the material, it is nice to know what I am doing and how it works. I honestly could care less about the probability of 3 girls being chosen for a position out of 8 with 7 males and 5 positions with 15 interviews.......

I appreciate the help.

## #17 Re: Help Me ! » probability » 2013-06-10 04:35:05

bobby,

thank you. I am quite proud of myself and could not have accomplished it without the help of this site. I love being able to understand what I am doing.

## #18 Re: Help Me ! » probability » 2013-06-10 04:27:26

Yes!!! Thanks so much for your help. I was able to calculate the probability of 9f, 7m for 5 positions with 16 finalists - the probability of at least 4f. I got it right!!! 3/13

## #19 Re: Help Me ! » probability » 2013-06-10 03:35:10

thank you so much!! I was doing 5!/14!(5-9)! and it was not coming out. This is difficult for me and I am sure that I will have more questions. Thanks again. At least now, I can figure the sample size.

## #20 Help Me ! » probability » 2013-06-10 02:44:25

mom
Replies: 19

I am completely lost and I need to know how to work out these problems. I have the following:

Suppose that 9 female and 5 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at random from the 14 finalists, what is the probability of selecting at least 4 females?

The experiment consists of selecting 5 finalists from 14. Since the order in which these 5 finalists are selected is immaterial, the samples are combinations of 14 finalists taken 5 at a time. Find the total number, N, of samples in the sample space.

N=C(14,5)
=2002

## #21 Re: Help Me ! » figuring mortgages » 2013-06-08 02:35:20

yeah, I think that I would prefer a fixed rate. After doing an amortized schedule for a variable rate, the balloon payment turned out pretty big. In fact it was bigger than the down payment.

## #22 Re: Help Me ! » figuring mortgages » 2013-06-06 10:13:06

johnnie,

thanks. I did figure it out. the 5/1 ARM is five year fixed rate and interest and payments change according to the market changes. The market declines, the interest and payments decline. The market increases - so do the payments and interest. The lender sets a floor and ceiling on the percentage and at the end of the term there is a balloon payment.

## #23 Help Me ! » figuring mortgages » 2013-06-06 04:56:46

mom
Replies: 6

I am writing a college paper on purchasing property. I have to include an excel spreadsheet on two different options. I understand how to do the fixed rate. It is easy. I am totally lost on the variable rate and I don't know how to begin to explain it. I have the following:

5/1 ARM (variable rate)
Interest rate    2.250%
APR        2.763%
Points        1.117
Closing costs    \$12,410.97
Monthly pymnt    \$1,529

7/1 ARM (variable rate)
Interest rate    2.750%
APR        2.897%
Points        0.825
Closing costs    \$11,282.99
Monthly pymnt    \$1,633

10/1 ARM (variable rate)
Interest rate    3.375%
APR        3.288%
Points        1.209
Closing costs    \$12,778.96
Monthly pymnt    \$1,768

Your results are based on the purchase of a home in ZIP code 22310, with an estimated purchase price of  \$499,990 and an estimated down payment of  \$99,998.

## #24 Re: Help Me ! » find the objective function » 2013-05-28 07:11:58

Thank you so much Bob!!

## #25 Re: Help Me ! » find the objective function » 2013-05-28 06:42:32

Bob,

thanks for all of your help. I got the correct answer to be 5x + 5y + 1800

It is correct. I double checked.