All Answers 359165 Answers

You are very welcome.  

Move onto your next question :)

A stick has a length of  units. The stick is then broken at two points, chosen at random. What is the probability that all three resulting pieces are shorter than  units?

First use contour mapping to graph the sample space.

Let the lengths be  x, y-x  and 5-y   

they add up to 5.

I know that  each of thise is between 0 and 5 units

0

0

0

x

y>x      (2)

0<5-y<5

-5<-y<0

0

Here is your sample space.  Every possible outcome is in there.

Now you have to graph the favourable sample space.  LOL you did not proof read your question!  

You want people to answer you.  You even tell an answerer that their answer is wrong.

Of course it is wrong, there is no question.  What a waste of our time. 

ok, I give up. I really can't comprehend the problem but I understand your explanation of your answer. Just as you say, maybe the choices are wrong. Thanks for your time.

My interpretation of the question

A coin is chosen and not returned BUT a dummy coin is thrown into the mix so that there will still be 10 coins to chose from

Then a second coint is chosen.  What is the probablility that this second coin is an old one.

I have answered this question correctly.

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Perhaps you are answering a different interpretation of the question ? I mean maybe you are trying to answering the intended question but it is not actually the question asked.

I think this maybe what you are trying to answer:

A coin purse has 3 OLD one-peso coins and 7 NEW one-peso coins. Suppose you randomly pick a coin and upon selecting, a dummy silver coin is placed on the purse as a placeholder for the coin you picked. If a second and third coin is picked from the coin purse, what is the probability that the second third coin is an OLD one peso? 

Does that sound right to you?  there are 3 picks so that is  2^3 = 8 outcomes.

I did look at this but I still got  27/100

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Unless you can think of a third interpretation of the question then I think none of those choices are correct.

Plus I do ot understand your working at all.   

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Perhaps you should move onto onother of you questions as I really believe that this one is a dud.  

Simplify

Hello Guest!

\(\large \frac{1}{\sqrt{2}+\frac{1}{\sqrt{8}+\sqrt{200}+\frac{1}{\sqrt{2}}}} \) = \(\frac{1}{\sqrt{2}+\frac{1}{\frac{\sqrt{8\cdot 2}+\sqrt{200\cdot 2}+1}{\sqrt{2}}}}\) 

= \(\frac{1}{\sqrt{2}+\frac{\sqrt{2}}{\sqrt{8\cdot 2}+\sqrt{200\cdot 2}+1}}\) = \(\frac{\sqrt{8\cdot 2}+\sqrt{200\cdot 2}+1}{(\sqrt{8\cdot 2}+\sqrt{200\cdot 2}+1)\cdot \sqrt{2}+\sqrt{2} }\) = \(\frac{4+20+1}{\sqrt{2}(16+20+1+1)}\) 

= \(\frac{25}{26\sqrt{2}}\color{blue}=\frac{25\sqrt{2}}{52}\)

  !

hi again, so I tried again and this is what I got

Assuming that an old coin was picked in the first, P(old, new) = P(2/10)(7/10)=7/50

Assuming that an either old or new was picked in the second and a dummy silver coin(maybe an old one or a new one)

P(old,new)= (1/10+1/10)(6/10+1/10) = 7/50

P=(7+7)/50=14/50 

is this reasonable? if this is not I'll just try asking my classmates how they get it lol

this is giving me a headache and I still got 2 more problems to solve.

btw thanks for your kindness.

Calculate the distance from earth to the sun in kilometers.

Hello Guest!

1 mile is equal to 1.609 kilometers.

\(93\ 000\ 000\ mi\times\dfrac{1.609\ km}{mi}=\color{blue}149\ 637\ 000\ km\)

  !

What prime is 4 greater than a perfect square and 9 less than the next perfect square?   

let P stand for the prime number

let x² stand for the first square

let (x+1)² stand for the next square   

x² + 4  =  P 

(x+1)² – 9  =  P  

therefore                          x² + 4  =  (x+1)² – 9  

                                        x² + 4  =  (x² + 2x + 1) – 9  

subtract x² from both sides     4  =  2x + 1 – 9  

combine like terms                12  =  2x

                                               x  =  6  

x² + 4  =  P  

36 + 4  =  40  

(x+1)² – 9  =  P 

49 – 9  =  40            P = 40 but 40 isn't prime so the problem has no answer 

_.

If A or B or E statements are true then C statement should be false

=> A/B/E broke it and C broke it 

Since above conclution contradicts 

A/B/E cannot be true.

If D statement is false, then A or C should be true. If A is true and C is false 

=> D broke it and C broke it (contradicts)

If A is false, the statement by A is false => D did not broke it which contradicts the conclusion that D is false (initial assumption)

Hence only option left is out

D is true => C is false => C broke it

I'm just kidding, but I still need help, how do I solve it algebraically?

OMG, that's so smart, why haven't I thought about that!!!!!!!

How do you do sin(18deg) without a calculator?  

Look it up in a book that has the trig tables.  

length of baseboard = Perimeter of a

around the room           Rectangle
Perimeter of a Rectangle = 2 Length + 2 Width 

                                or     = 2 (L + W)

*Make it an improper fraction then use

their LCD to be a common fraction to be added

L = 7 1/17 = \({(17)(7)+1\over 17} = {120\over 17} = {(34 / 17) * 120\over 34} = {240\over 34}\)

W = 5 1/2 = \({5(2)+1\over 2} = {11\over 12} = {(34 / 2) * 11\over 34} = {187 \over 34}\)

Least Common Denominator (LCD)

or Least Common Multiple (LCM) of 17 and 2:

17: 17(1)=17, 17(2)=34

2: 2(17) = 34

P = 2(240/34 + 187/34) = 2¹(427/34₁₇) = 427/17 feet

Cost = (427/17 feet) ($3.49/ft) = $87.66

Thanks Alan,

I still don't  get it but that doesn't matter.  (I have a calculator :/)

Hopefully the shadow will understand.

You are very welcome.

I think it is a classroom competition question because it has been asked by about a dozen different guests in a matter of about an hour.

People are obviously cheating. It was most likely a fun exercise and there was some small reward for the first on to present the answer.  Or maybe it was in a bigger competition. 

No matter, there is no doubt in my mind that ALL of the askers are cheating.

Denominator cannot equal 0

(x^2+bx+88)       soooo      b^2 - 4(88)    >=0

                                         b^2 = 352

                                         b=+-18.76             largest integer 18

hi, thanks for responding.

Please show hotw you got 7/50.  If you do not show working I have  no ability to discuss it with you.

I am sticking by my solution. I do not believe any of the given answers are correct.

If you are given a worked answers that gives a different answer from me, please present it here.

I answered this one not very long ago. You could go look for my solution..

2\log{(a -2b)} = \log{a} + \log{b}.

\(2\log{(a -2b)} = \log{a} + \log{b}\\ \log{(a -2b)^2} = \log{(a*b)}\\ (a-2b)^2=ab\)

etc

F * [F + 2D]==9

F==1 - this is the first term

D ==4 - this is the common difference

Your sequence looks like this:

1, 5, 9, 13, 17, 21, 25........and so on.

The 4th term ==9 + 4 ==13

A is true? Then D is also True. X

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

D is T? Then A, B, E, C are F.

     D, E, B didn't break it.

      C break it

So D is T, C break it ✓

Only D is true, C break it

Therefore, C broke the vase

If we add all the equations, we get \(10a + 10b + 10c + 10d = 0\).

We then divide by 10 to get \(a + b + c + d = \boxed{0}.\)

(a) Let w = a + bi and z = c + di.  The rest is expanding.

(b) Let w = a + bi and z = c + di.  Then

\(\dfrac{w + z}{1 + wz} = \dfrac{a + c + bi + di}{1 + (a + bi)(c + di)}\)

To express this in rectangular form, we can multiply the numerator and denominator by the conjugate:

\(\dfrac{a + c + bi + di}{1 + (a + bi)(c + di)} = \dfrac{(a + c + bi + di)((1 - (a + bi)(c + di))}{(1 + (a + bi)(c + di))(1 - (a + bi)(c + di))}\)

The denominator simplifies to (1 - (a^2 + b^2)(c^2 + d^2)), which is real.  The numerator simplifies to a^2 - b^2 + c^2 - d^2, which is also real.  Therefore, the complex number (z + w)/(zw + 1) is real.