All Answers 358108 Answers

ur welcome Alan!and its ok to make these kind of mistakes , it happens with everyone ! nothing too serius !but for a tip while typing on LaTex u shuld move it at the corner of the wrting center so the ques and other things are easily visible!

u know reinout i could understand only the last line of both of ur notes but it was interesting reading it !i didnt know that these things are provided in the lunchbox now a days !lol!

 gift this from me to Beth for the prom! take care of my dress Beth !

I think your math book might have made a mistake. 2 * (-3) = -6

Because

positive*positive = positive

negative*positive = negative

positive*negative = negative

negative*negative = positive

Maybe if there's more to the question than 2*(-3) you should check the rest of the question 

That wasn't the point Alan.

The point is that the card has already been chosen.  It has 1 red side. therefore

P(2whites sides)=p(2black sides )=p(2purple sides)=0

p(2red sides)=P(one red and one white)=0.5

How can you say differently?????

Please try and explain to me what is wrong with my logic. (or lack thereof)

Okay Alan 

I only just read your answer properly - Now I have something to think about - thank you.

Yea I get it now I think.

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Ok Ok I've changed my mind - I'm wrong - You're right - who is to be Sir Lancelot, head knight of camelot?

Is it Alan? Is it Bertie?  Let the jousting begin!

I'm afraid your analogy is misleading you Melody!  There is only one way of "choosing" a boy.  There are two ways of choosing a red side from a card with two red sides.

I just lost sight of the fact that the last term was negative while concentrating on all the LaTeX stuff!  Thanks for correcting this rosala.

so well solve it like this ! alan has done it that way but i find it easier doing it directly !so its like this 

8/25 + 5/7 - 1/2

=112 + 250 -175/350

=362 - 175/350

=187/350 = 0.5342857142857143

so thats it but alan i dont understand that why did u plus the three nos when the question was like this  8/25 + 5/7 - 1/2  !

Monday 16/6/14

Thank you NinjaDevo

I'm sticking to my answer too.  I believe that is like saying that the chance of a child being a boy is 50%.  If you already have two boys and there is another baby on the way what is the chance of ending up with 3 boys.  it is not 0.5³   It is 0.5

Well the card is chosen.  It is definitly not the fully white one.  that is no longer a possibility so the probability that the other side will be white is 50%. the probablity that it is red is 50%.

I know Alan and Bertie are superior mathematicians but I am STILL sticking with my answer!

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NO I'M NOT - I'm Lady Guinevere - My mind can change as doth the vector of the breeze!

The maker of the glue shall be banished from the land of camelot for it has become Unstuck!

Express each term with a common denominator.  Easiest way to find a common denominator is to multiply all three denominators together.

$$$$\frac{8}{25}*\frac{25*7*2}{25*7*2}+\frac{5}{7}*\frac{25*7*2}{25*7*2}+\frac{1}{2}*\frac{25*7*2}{25*7*2}$$\\\\
$$\frac{8*7*2}{25*7*2}+\frac{5*25*2}{25*7*2}+\frac{1*25*7}{25*7*2}$$\\\\
$$\frac{8*7*2+5*25*2+1*25*7}{25*7*2}$$\\\\
$$\frac{112+250+175}{350}$$\\\\
$$\frac{537}{350}$$ 

$${\frac{{\mathtt{537}}}{{\mathtt{350}}}} = {\mathtt{1.534\: \!285\: \!714\: \!285\: \!714\: \!3}}$$

or  

 $$1\frac{187}{350}$$

Edit:  I put a "+" instead of a "-" for the last term, so the final answer is incorrect!  See rosala's post for the correct answer. 

I'm with Alan, the probability is 2/3.

There are six possible outcomes. I can draw any one of the three cards and I can then see just one of its sides. Imagine a letter on each side of each card, so, card 1 side A, card 1 side B, card 2 side A etc. If I were to draw card number 1, I could see either A or I could  see B, that's two possibilities, not one. Same for the other two cards.

Of the six possible outcomes, three will be red and three will be white.

Of the three possible reds, two will have red the other side one will have white. (We have two possible ways in which the other side is red, one on which the other side is white.) So, the odds are 2 - 1 on the other side being red, a probability of 2/3.

Alan's first post shows this quite clearly.

ur welcome melody !

zegroes and CPhill u all answered really very good but i also wanna answer this question so here we go :

5/4 x 9/10

=1/4 x 9/2   (here we have cut 5 and 10)

=9/8 = 1 1/8  (this is the mixed form of 9/8 , i just did for just if u need it ) =1.125 (thats the decimal form )

i gave u the answer in three forms  becoz i thought u could use the one u require or ur teacher teaches u !so i just hope this helps !

I'm afraid you're wrong guys!  This is a well known problem in recreational maths.  

P(A and B) = Probability of two sides red = 1/3 (there are 3 cards, one of which has two red sides).  

P(A) = Probability of one side red = 1/2 (there are 6 sides, 3 of which are red).

P(A and B)/P(A) = (1/3)/(1/2) = 2/3.

Thank you Heureka.

I got that. Back to kindergarten for me!

Hi Melody,

0 + 2 = 2    n o t  0

We need to convert the 17 minutes into a decimal.....since there are 60 minutes in one degree, we have

17 minutes *  1 degree/60 minutes  =  17/60 degree  =  .2833333333333333 degrees   (Note how the minutes "cross-cancelled" leaving just degrees !!)

So...we are evaluating...

sin (34.2833333333333333°) ≈ .5633

One terabyte = 10¹² bytes....so........

5 million * 10¹² =

[5 *10⁶ * 10¹² ] =  5 * 10¹⁸ bytes =   5,000,000,000,000,000,000 bytes

Or, using the US definition of this.....five quintillion bytes-......

Well, you have two different questions...the one in the title and the one in the body of your question.

Title question: 162 x 171 = 27,702

Body question: 170 x 122 = 20,740

I rounded up to $62.52, but a penny isn't worth quibbling about!!

I'm always impressed with your ability to use LaTex, heureka !!!   ......I wish I could learn that, but I'm probably too old !!!!

I'm an idiot, Jedithious,  because I CAN'T READ !!!   Ignore my stupid answrs to 3b  and 3c !!!!

$$\\A=\left(
\begin{array}{cc}1 & 2 \\3 & 4
\end{array}
\right)\qquad
B=\left(
\begin{array}{cc}2 & 6 \\-1 & 0
\end{array}
\right)$$

$$\boxed{AB+B=(A+\textcolor[rgb]{0,0,1}{I})B}$$

$$\mbox{Identity Matrix }\textcolor[rgb]{0,0,1}{I=\left(
\begin{array}{cc}1 & 0 \\0 & 1
\end{array}
\right)}\\$$

$$\mbox{A+\textcolor[rgb]{0,0,1}{I}} =\left(
\begin{array}{cc}1 & 2 \\3 & 4
\end{array}
\right)
+
\textcolor[rgb]{0,0,1}{\left(
\begin{array}{cc}1 & 0 \\0 & 1
\end{array}
\right)}
}
=
\left(
\begin{array}{cc}1+\textcolor[rgb]{0,0,1}{1} & 2+\textcolor[rgb]{0,0,1}{0} \\3+\textcolor[rgb]{0,0,1}{0} & 4+\textcolor[rgb]{0,0,1}{1}
\end{array}
\right)
=
\left(
\begin{array}{cc}2 & 2 \\3 & 5
\end{array}
\right)$$

$$\mbox{(A+\textcolor[rgb]{0,0,1}{I})B}=\left(
\begin{array}{cc}2 & 2 \\3 & 5
\end{array}
\right)
}
\left(
\begin{array}{cc}2 & 6 \\-1 & 0
\end{array}
\right)
=
\left(
\begin{array}{cc}\textcolor[rgb]{1,0,0}{2} & 12 \\1 & 18
\end{array}
\right)$$

$$\boxed{AB+B=
\left(
\begin{array}{cc}\textcolor[rgb]{1,0,0}{2} & 12 \\1 & 18
\end{array}
\right)
}$$

First add the quarts = 2+ 1 = 3....(if this answer would have been 4 or greater we would have converted it to gallons and quarts....for example: 3qts + 3 qts = 6 qts = 1 gallon and 2 qts.

Now, add the gallons = 5 + 6 =  11

So, we have 11 gallons and 3 qts.