Bxtterman

Thanks! I'll try to answer 1-2 questions per day.

Hmm... Not math, but interesting!

I'm not so good at these types of problems, please forgive me if this isn't the answer that you wanted (I messed up super bad on another problem!)

Here we go:

I believe that the sample is biased because it does not represent the population - The City did not get random coaches, so it isn't random.

I also believe that the question is not biased - because both water and the sports drink are given equal opportunity. UNLESS you count in the order, where drinking water is presented first as an option, and therefore has a slight psychological effect upon the participant's mind. :)

My own thoughts -- probabaly not professional. :)

My fault! Sorry! :(

My Bad! I was too carred away by the LaTeX

Thanks

If LaTeX is the only way someone can type on a computer, I find that very weird. Additionally, you should be able to convert it into somewhat readable form. Also, there is a LaTeX function in the Web2.0Calc editor. I don't want to argue about this, just wanted to let you know that you should create a better story next time. Thanks.

Hi, thanks for posting. We will use heron's formula for this, because I assume if you posted on this forum you would have already had googled a triangle area calculator.

Heron's Formula is:

\(A=\sqrt {s(s-a)(s-b)(s-c)}\)

\(s\): The Semiperimeter of a triangle (Half of the perimeter)

\(a,b,c\) = The sides of the triangle.

First Question:

\(A=\sqrt {1.5(1.5-1)(1.5-1)(1.5-1)}\\A=\sqrt {1.5(.5^3)}\\A=\sqrt {0.1875}\\A \approx \boxed{0.433}\)

Second Question:

\(A=\sqrt {50(50-49)(50-33)(50-18)}\\A=\sqrt {50(544)}\\A=\sqrt {27200}\\A \approx \boxed{164.92}\)

Specify if you want a certain formula used, thanks.

Thank you for trying before you post on this forum. I appreciate it. (I am really tired of all the people just being lazy and not trying at all and posting on here).

As for your question.

To find angle \(Q\) and angle \(P\), you need to first solve for \(y\) . We first solve for the unknown angle \(\angle QRP\). We see that this angle and the angle \(\angle QRS\) add up to 180º (Straight Line). As you already did, 180º-150º is 27º.

We know that the sum of the angles in a triangle add up to 180º, so we can now make a linear equation to solve for \(y\). The Equation is as follows:

\((3y+5)+(2y-7)+27=180\)

We then solve for y:

\((3y+5)+(2y-7)+27=180 \)

\( 3y+5+2y-7+27=180\) -- Remove Parentheses

\(5y+25=180\) -- Combine like terms

\(5y=155\) -- Isolate the Variable

\(y=31\)

We can now solve for each of the angles.

\(m\angle Q= 3(31)+5\\m\angle Q= 93+5\)

\(m\angle Q= \) \(\boxed{98}\)

\(m\angle P=2(31)-7\\m\angle P=62-7\\m\angle P=\boxed{55}\)

\(m\angle QRP=\boxed{27}\) (We already solved this.)

This was my first time using LaTeX, please forgive me. :)

AoPS I assume? Or some online class for math? My tips for you. These classes are expensive, and please try to do your homework and at least try to solve the problems before you ask here. If you did indeed attempt to solve these problems, please show your steps to solve. Furthermore, your effort on the post is super minimal, and you don't even format it correctly. At least take the format out of LaTeX and into readable form before you post. We are here to help you learn, not provide answers. If you really don't want to learn at your class/course, please just don't waste your time or sign up for it. Just a message and reminder from me.