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0.4 x 7 mph =2.8 miles - distance run.
Let the required speed =S
[2.8 + 0.8S] / [0.4+0.8] =5, solve for S
S =4 mph - must walk to attain an average of 5 mph, or:
6 miles / 1.2 hours =5 mph

1 - Find the volume of the pool in cubic feet by multiplying 3  x  1.5  x  3.5

2- Once you find the result in (1) above, then multiply it by 7.5 gallons, which will give you the answer.

This consists of the sum of : 5C0 + 6C1 + 7C2 + 8C4 + 9C5 =1 + 6 + 21 + 56 + 126 =210 numbers.

Since there is an increase in manpower/hours of: [8 x 4] / [6 x 5] = 16/15, therefore, there must be decrease in days by the same pecentage. Or 16 / [16 / 15] =16 x 15/16 =15 days

On a road the number of cars and the number of lorries are in the ratio 10 : 1 the number of lorries and the number of vans are in the ratio 7:10 There are 50 vans on the road. How many cars are there on the road?

7 / 10 = T / 50, solve for T

T = 35 trucks (lorries?)

Since the ratio of cars to trucks(lorries) = 10: 1, therefore:

There must be:35 x 10 = 350 cars 0n the road.

What is the smallest positive five-digit integer, with all different digits, that is divisible by each of its non-zero digits? (The number itself may have 0 as a digit.) 

Does the problem allow the 5-digit integer to begin with a zero? 

If so, then submitted for your approval is the number 01236 

The problem does not explicity prohibit the leading zero. 

If you make the assumption that a leading zero is not allowed,

and I make the assumption that a leading zero is allowed, 

then who says my assumption is less valid than yours?

_.

(a) In order to solve this problem we will introduce new variables u and v, where u = 4t - 4 and v = -2t + 7.  Notice that as t gets very large, u = 4t - 4 looks like 4t, and v = -2t + 7 looks like -2t.  We can formalize this by saying that as t gets very large, v/u = (-2t + 7)/(4t - 4).  Divide both sides by t, to get v/u = (-2 + 7/t)/(4 - 4/t).  As t gets large, 7/t and 4/t are small, so u/v becomes (-2 + 7/t)/(4 - 4/t) = -1/2.  This tells us that we should look for a line of slope -1/2.

If y/x = (-2t + 7)/(4t - 4), then y(4t - 4) = x(-2t + 7), so 4yt - 4y = -2xt + 7x.  Then 4y + 7x = 4yt + 2xt = t(4y + 2x).  Since this is true for every value of t, we must have 4y + 7x = 0 and 4y + 2x = 0.  When we solve this system, we get that x = y = 0, so that tells us that our line passes through the origin.  And we know that the slope of the line is -1/2, so the equation of our line is y = -x/2.  But that is only the base line - we can take our line and shift it in the x-direction, or y-direction, until it matches what we want.  Here, the problem tells us that our line passes through the point (4,3), so we can shift it 3 units up, to get y = -x/2 + 3.  We can then shift it four units in the x-direction, to get y = -(x + 4)/2 + 3 = -x/2 + 5.  So this is the equation of the line.

(b) We can use the same argument as in part (a).  Let u = 4t - 4 and v = -2t + 7.  As we said in part (a), as t gets very large, u/v gets close to -1/2.  So we can find a value of t so that u/v is close to -1/2.  If y = -x/2 + 5, then y/x is close to -1/2.  So choose the value of t so that y/x = u/v.  Then y/x = (-2t + 7)/(4t - 4).  This leads to the same equation 4y + 7x = 4yt + 2xt = t(4y + 2x).  Then again, 4y + 7x = 0 and 4y + 2x = 0, so x = y = 0.  This gives us a base line of y/x = -1/2, or y = -x/2.  We can then shift five units up, because we want the line to pass through (0,5), to get the line y = -x/2 + 5.  So the line of y = -x/2 + 5 is the graph of G.

is 1/5*30=6 

Yes, it is.  Matbe this is more easily visualized with the implied denominator written explicitly. 

30 has an implied, or understood, denominator of 1.  So if you write your problem thus:

1       30      30 

—  x  —  =   —  =  6 

5        1        5 

_.

1 week = 7 days = 24 hours.

7 x 24 =168 hours in 1 week

N =5,000 x [1 + 0.0008]^168

N =5,000 x [1.0008]^168

N =5,000 x      1.143788809294......

N =~5,719 cells after 1 week.

Maybe, you could verify it this way:

The probability of having AT LEAST 2 of a kind =16 / 216. Since this applies to all 6 numbers, then the probability is:16 x 6 = 96 / 216. Now, this is the probability of at least 2 of a kind appearing. His question is:

"What is the probability that, in a roll of 3 fair dice, the same number is NOT rolled twice?". Then my understanding is that we want the opposite and subtract: [216 - 96] / 216 = 120 / 216 =5 / 9

72 gm Fe / 9 gm Zn   *  144 gm Zn   =

72/9    *  144    gm Fe   =      ???        I think you can do it now.....

x = miles = 150 miles

g(x) = gallons left in tank  (what you are looking for in the Q)

g(x) =  - .04 (150) + 10       I think you can do it now.....

For question 1 I would label the dice A, B and C.

Whatever number A shows, the probability that B shows something different is 5/6. The probability that C then shows a different number from both A and B is 4/6.  Hence overall probability is (5/6)*(4/6) = 20/36 = 5/9.

To complete a task in 16 days a company needs 5 people each working in 6 hours per day The company decides to have 8 people each working in 4 hours per day Assuming that each person works at the same rate,how many days will the task take to complete

Ok thanks!

And also, Guest, I'm not sure why you deleted your answer, but if you were trying to hide it for some reason, I still saw your answer.

I was trying to solve this problem without a calculator, and I was wondering if there was a trick, so I found this:

This sum consists of terms from the 13th row of Pascal's Triangle. We know the sum of the entire row is

Our sum consists only of the first half of those terms. Since Pascal's Triangle is symmetric, the sum of the terms in the first half of the row is equal to half the sum of the whole row's terms:

Why did you get a point for that?

ok it is combination maths.

You can do it on your calculator.

13 nCr 0 + 13 nCr 1 + etc

I'm pretty sure it is 

Are they combination maths or vectors?

Yes, I have a pointer.

1 post = 1 question.

Yes, I already apologized.

I am also sorry if I caused you to waste your time with rechecking.

Thank you for showing us what you have attempted. 

That is what you should do EVERY time.

I for one cannot think straight with all those questions bombarding me.

One post is for one question.  I must have said that 1000 times.

But I will take a look at question 1

1) What is the probability that, in a roll of 3 fair dice, the same number is not rolled twice?

I listed out all the posssible combinations that could have happen with 1. I got 20. I multiplied that by 6 and got 120. Then I know that 6 cubed is 216... so the I thought the probability must be 120/216 = 15/27.

You listed all the possibilities .....  Where is your list?  Do you mean a list where one only happens once?

one of your combinations would be 1,2,3,

Now  if you do the same thing for the number 2 and add them together then this combination will appear again.

Hence I expect you have double counted. 

See I can teach you if you show me where you are at!  Well maybe anyway.

With 3 rolls there is a 6^3 = 216 possible combination,

111       1 way

112        3ways

113         3ways

114         3ways

115         3ways

116         3ways

So that is 16 ways     (maybe that is what you did and one of us counted wrong?)

the same would  be for the one 5 numbers so

16*6= 96

so that appears to be   96/216 = 4/9

Edit:

This is the probability that a number will repeat.

The probability that a number will not repeat is 1-4/9 = 5/9

Now the idea is that if possible you see if my answer is correct.

(If you know what the answer should be your should already have told us)

If the answer is wrong then you politely and clearly explain why you think this. 

(eg I put the answer in the answer box and the computer said it was wrong)

If the answer is correct then you try to see what I have done right and what you did wrong.

If you have questions then you ask for further explanation.

After you have digested all this and learned all that you can learn,

Then

You attempt the next question again by yourself and if you can not do it you present the next question and what you have tried in a completely new post. 

-----------------------------------------------

Now digest all I have told you .

You could start by deleting questions 2 to 6 from this post.

One post = one question.