Let's actually find the line of best fit...

m=(nΣyx-ΣyΣx)/(nΣx^2-ΣxΣx)

m=(11*836-130*55)/(11*385-3025)

m=2046/1210

m=93/55

b=(Σy-93Σx/55)/n

b=(55Σy-93Σx)/(55n)

b=(7150-5115)/(55*11)

b=185/55, so the line of best fit is:

y=(93x+185)/55

A) The approximate y-intercept (the value of y when x=0) is 185/55≈3.36.

Which means that those who do not practice at all will win about 3.36 times

B) y(13)=(93x+185)/55

y(13)≈25.34

So after 13 months of practice one would expect to win about 25.34 times.

3 0

3 years ago

Write each expression as an algebraic (nontrigonometric) expression in u, u > 0.

max2010maxim [7]

Answer:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

Step-by-step explanation:

We want to write the trignometric expression:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0$

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

$\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)$

Take the secant of both sides:

$\displaystyle \sec(\theta)=\frac{u}{10}$

Since secant is the ratio of the hypotenuse side to the adjacent side, this means that the opposite side is:

$\displaystyle o=\sqrt{u^2-10^2}=\sqrt{u^2-100}$

By substitutition:

$\displaystyle= \sin(2\theta)$

Using an double-angle identity:

$=2\sin(\theta)\cos(\theta)$

We know that the opposite side is √(u² -100), the adjacent side is 10, and the hypotenuse is u. Therefore:

$\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)$

Simplify. Therefore:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

4 0

2 years ago