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Manny H.
Statistics grad with a love for math and teaching.
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Calculus
TutorMe
Question:

Solve this integral, $(\int_{}^{} \frac{3x^2}{(x^3+3)^2}dx$)

Manny H.

We can solve this integral using the u-substitution method. Let $( u = x^3+3$) Find the derivative of $$u$$ and solve for $$dx$$ to plug in, $(du = 3x^2dx$) $(dx = \frac{du}{3x^2}$) Plug $$u$$ and $$dx$$ into original equation, $(\int_{}^{} \frac{3x^2}{(x^3+3)^2}dx$) becomes $(\int_{}^{} \frac{3x^2}{u^2}\frac{du}{3x^2}$) $(\int_{}^{} \frac{du}{u^2}$) $(\int_{}^{} \frac{-1}{u} + c$) Plug the value for u back in and your answer is: $( \frac{-1}{(x^3 + 3)} + c$)

Statistics
TutorMe
Question:

Probability: We want to randomly select 5 dogs from a group of 10 dogs. Of those 10 dogs, 3 are female. What is the probability that we get all 3 female dogs in our selection?

Manny H.

This probability problem can be easily represented by the Hypergeometric distribution because we are drawing without replacement and we can view picking a female as a success. So, the probability distribution can be represented by the expression $(\frac{(_KC_k)(_{N-K}C_{n-k})}{_NC_n}$) where $(\text{k = number of successes n = number of draws}$) $(\text{N = population size K = the total number of possible successes}$) In this case K and k both equal 3. N is 10 and n is 5. Plugging in we get, $(\frac{(_3C_3)(_{10-3}C_{5-3})}{_{10}C_5}$) Solving we get: Probability of 3 female dogs in selection = 0.083

Algebra
TutorMe
Question:

Solve the following equation for x, $( |-4x + 2| = 1$)

Manny H.

To solve this problem we need to understand how to break down absolute value equations into 2 separate equations. For example, given $$|x + a| = n$$, we can break it down as follows: $(x + a = n$) $(and$) $(-(x + a) = n$) then solve for 2 different x values. So for this problem, we will have $$-4x + 2 = 1$$ and $$-(-4x +2) = 1$$. Solving: $( -4x + 2 = 1$) $( -4x = -1$) $( x = 1/4$) and $( -(-4x + 2) = 1$) $( 4x - 2 = 1$) $( 4x = 3$) $( x = 3/4$) So, our solutions are $$x = 1/4, 3/4$$

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