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Brittany A.
Math Tutor for Three Years
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Basic Math
TutorMe
Question:

Triangle ABC is a right triangle. The hypotenuse of the triangle is 13 inches, and one of the legs is 5 inches long.

Brittany A.

We can use the pythagorean theorem to solve this. $$c^2=a^2+b^2$$. Where c is the hypotenuse and a and b are the legs. We plug in c=13, and a=5. $$13^2=5^2+b^2$$ which simplifies to $$169=25+b^2$$ We can subtract 25 from both sides to get $$169-25=25-25+b^2$$, $$144=b^2$$, and taking the square root, we get $$b=12$$. Therefore the other leg is 12 inches.

Statistics
TutorMe
Question:

A poll for an election with two candidates is conducted with a simple random sample of 100 people from a population of one million people. Candidate A got 45% percent and Candidates B got 55% percent. What is the standard error of means for this scenario?

Brittany A.

The poll is a simple random sample so we know we can use the standard error of means. The formula for standard error of means is: $$SE=\sqrt {\frac{a(1-a)}{n}}$$. Let a=.45 and n=100 $$SE=\sqrt {\frac{.45(1-.45)}{100}}=.0497$$ The standard error of means in this case is 0.0497.

Algebra
TutorMe
Question:

How do I solve the equation $$x^2+6x+13=5$$ with quadratic formula?

Brittany A.

The first step is to subtract 5 from both sides so we can solve it. $$x^2+6x+13-5=5-5$$ $$x^2+6x+8=0$$ The equation $$x^2+6x+8-0$$. The quadratic formula is $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ for an equation $$ax^2+bx+c=0$$ In this case a=1, b=6, c=8. We plug in the values to the equation. $$x=\frac{-6\pm\sqrt{6^2-4(1)(8)}}{2(1)}$$. We can do some simple addition, subtract and multiplication to simplify the equation. $$x=\frac{-6\pm\sqrt{36-32}}{2}=\frac{-6\pm\sqrt{4}}{2(1)}=\frac{-6\pm 2}{2}=3 \pm 1$$. The solutions are x=3+1, x=3-1, which simplify to be x=4, x=2.

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