Improve Your Algebra Skills with Practice Problems

Confusion on simple probability question
Suppose you take a ball at random without knowing which urn you took it from. Here are two kinds of urns, one with three red balls and one blue, and one with three blue balls and one red. Both urns are equally likely. The ball turns out to be blue. What is the probability of getting two blue balls without knowing which bin they came from?
Answer: 5/16
I'm confused on how this answer is derived. My reasoning was that there are 8 balls and 4 are blue. The chance of drawing a blue ball the first time is $\frac{1}{2}$ and the second time is still $\frac{1}{2}$ (assuming replacement). $\frac{1}{2}\ast <!--\; \ast \; -->\frac{1}{2}=\frac{1}{4}$

Round $604.2978$ to the hundredths place.

Find:
$$\frac{2}{3}=\frac{4}{x}$$

In partial fractions, why must the degree of the numerator be lower than the denominator?
Specifically, it must be one degree lower. But why must it be smaller?

Solve:
$$\frac{1}{\sqrt[3]{4x}}$$

How to prove that $\frac{a^2+b^2-<!--\; -\; -->2}{a+b-<!--\; -\; -->2}<1+b$ if $a<b$ and $a,b>1$?

How many real numbers satisfy $x+x^{-<!--\; -\; -->1}$ so that it is a Whole number.
I came across this in a discrete mathematics book, and i wonder if my solution is correct.
From looking at $y=x+x^{-<!--\; -\; -->1}$
I see only 2 possibilities, x ={1;-1;}
Was this just so easy or am I missing something?

What is the recommended method to simplify trigonometric expressions like $\frac{\sin <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->\sin <!--\; \; -->\mathrm{\beta <!--\; \beta \; -->}}{\cos <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->\cos <!--\; \; -->\mathrm{\beta <!--\; \beta \; -->}}$?
I tried to manipulate them using double and sum but none of them seemed to help.

Let $(x,y,z)\in <!--\; \in \; -->\mathbb{R}$, prove that $(\frac{2x-<!--\; -\; -->y}{x-<!--\; -\; -->y}{)}^2+(\frac{2y-<!--\; -\; -->z}{y-<!--\; -\; -->z}{)}^2+(\frac{2z-<!--\; -\; -->x}{z-<!--\; -\; -->x}{)}^2⩾<!--\; ⩾\; -->5$

Body temperature is usually maintained between ${36.5}^{\circ <!--\; \circ \; -->}C$ and ${37.5}^{\circ <!--\; \circ \; -->}C$, inclusive. Determine the corresponding range of temperatures in Fahrenheit. Use the relationship between degrees Celsius C and degrees Fahrenheit
The temperature of the body in Fahrenheit at 36.5 degree Celsius and 99.5 degree Celsius: $F:C=\frac{5}{9}(F-<!--\; -\; -->32)$

1 yard = 3 feet 1 foot = 12 inches 1 inch = 2.54 cm Convert 2 yards to centimeters (round to the nearest whole centimeter).
A) 30 centimeters
B) 36 centimeters
C) 91 centimeters
D) 183 centimeters

Find an equation of the line containing the given pair of points. Write your final answer as a linear function in slope-intercept form.
(2, 0) and (7, 9)

$$P=f(t)$$ gives the size of a population that begins with 19,000 members and grows at a continuous annual rate of 1.71%.
Find a formula for the population.

For a recent year, there were 10 named tropical storms and 6 hurricanes in the Atlantic. Write a ratio of the number of hurricanes to the total number of named storms.

Write an equation for a linear function whose graph has given characteristics.
Passes through (5,0), parallel to the graph of $g(x)=-<!--\; -\; -->\frac{4}{5}x-<!--\; -\; -->6$

Let $a_{n+1}=\frac{2a_n+1}{a_n+1}$ .For $a_1$=1 can we show that $a_n$<2

Prove this inequality $\sqrt[3]{(a+b+c)(b+c+d)}\ge <!--\; \ge \; -->\sqrt[3]{ab}+\sqrt[3]{cd}$
Let $a,b,c,d>0$
Idea: Hence, we need to prove that
$(a+b+c)(b+c+d)\ge <!--\; \ge \; -->ab+cd+3\sqrt[3]{(ab{)}^{2}cd}+3\sqrt[3]{(cd{)}^2(ab)}$
$⟺<!--\; ⟺\; -->ac+bc+bd+ad+b^2+c^2\ge <!--\; \ge \; -->3\sqrt[3]{(ab{)}^{2}cd}+3\sqrt[3]{(cd{)}^2(ab)}$
I attempted a following proof by AM-GM inequality,But I am not able to solve the upper part, this inequality

How to prove $\sqrt{1000}<x<1000$?
I have been given that
$x=\frac{2}{1}\times <!--\; \times \; -->\frac{4}{3}\times <!--\; \times \; -->\frac{6}{5}\times <!--\; \times \; -->\frac{8}{7}\times <!--\; \times \; -->\cdots <!--\; \cdots \; -->\times <!--\; \times \; -->\frac{996}{995}\times <!--\; \times \; -->\frac{998}{997}\times <!--\; \times \; -->\frac{1000}{999}$
How can I prove that $\sqrt{1000}<x<1000$?

Show that $\underset{x\to <!--\; \to \; -->0}{lim}\frac{x(1+\cos <!--\; \; -->(x))-<!--\; -\; -->2\tan <!--\; \; -->(x)}{2x-<!--\; -\; -->\sin <!--\; \; -->(x)-<!--\; -\; -->\tan <!--\; \; -->(x)}=7$?

One ounce is equal to 28.35 grams. convert 16 ounces to grams round your answe to the nearest tenth