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Option (A) evaluates to $10.$

Option (B) evaluates to $0 + 1 + 7 = 8.$

Option (C) evalutes to $2 + 0 + 7 = 9.$

Option (D) evaluates to $2 + 0 + 7 = 9.$

Option (E) evalutes to $0.$

Thus, A is the correct answer.

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If $36$ votes is $30\%$ of the total votes, then $10\%$ of the total votes is $12$ votes. The number of total votes would then be $10 \cdot 12 = 120.$

Thus, E is the correct answer.

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This expression can be reduced as follows: $$\begin{align*} \sqrt{16\sqrt{8\sqrt{4}}} &= \sqrt{16\sqrt{16}} \\ &= \sqrt{64}\\ &= 8 \end{align*}$$

Thus, C is the correct answer.

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We can approximate the product as $$\begin{align*} (3 \cdot 10^{-4})(8 \cdot 10^6) &= 24 \cdot 10^2 \\ &= 2400. \end{align*}$$

Thus, D is the correct answer.

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Looking at the denominator independently, $1 + 2 + 3 + \cdots + 8 = 36,$ so the desired answer is $$\begin{gather*} \dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \\ 4 \cdot 5 \cdot 7 \cdot 8 = 1120. \end{gather*}$$

Thus, B is the correct answer.

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We can let the three angles be equal to $3x, 3x,$ and $4x.$ Then we know that their sum equals $180.$ From this we can set $3x + 3x + 4x = 180,$ and solving this, we get $10x = 180$ and $x = 18.$

The largest angle is $4x = 72.$

Thus, D is the correct answer.

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We can let $Z$ have the form $abcabc.$ Then, we get that $$\begin{align*} Z = 1001 \cdot abc = 7 \cdot 11 \cdot 13 \cdot abc. \end{align*}$$ This means that $11$ is a factor of $Z.$

Thus, A is the correct answer.

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If a number is even, it is divisible by $2.$ This means that the house number is either divisible by $2$ or by $7.$ This rules out the possibility that the house number is prime. This means that the house number is divisible by $2$ and by $7,$ and one of its digits is $9.$

The divisibility conditions show that the house number is divisble by $14,$ so if we look at all the two-digit multiples of $14,$ the only one with a digit of $9$ is $98.$ Therefore, units digit is $8.$

Thus, D is the correct answer.

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If the number of marbles is divisble by both $3$ and $4,$ then the number must be divisible by $12.$ If we test $12,$ we get that there are $4$ blue marbles and $3$ red marbles. This leaves a maximum of $5$ green marbles, which is not possible.

If there are $24$ marbles, then there are $8$ blue marbles and $6$ red marbles. To find the number of yellow marbles, we get $24 - 8 - 6 - 6 = 4.$

Thus, D is the correct answer.

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The number of ways to choose $3$ cards from $5$ is ${5 \choose 3} = 10.$ If $4$ is the largest value selected, then the other two cards have to be chosen from ${1, 2, 3}.$ There are ${3 \choose 2} = 3$ ways to do this. The probability is then $\dfrac{3}{10}.$

Thus, C is the correct answer.

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$37$ tiles on both diagonals imply that there are $19$ tiles on each diagonal, since one tile overlaps in the middle. The total number of tiles would then be $19^2 = 361$ since the number of tiles in each row is equal to the number of tiles in one diagonal.

Thus, C is the correct answer.

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If a number leaves a remainder of $1$ when divided by $4, 5, \text{ and } 6,$ then it is one more than the least common multiple of these numbers. The least common multiple is $60,$ so the smallest such positive integer is $60 + 1 = 61.$

Thus, D is the correct answer.

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Since there are no ties, the number of wins has to equal the number of losses. The number of losses is $2 + 3 + 3 = 8,$ so the number of games Kyler won is $8 - 4 - 3 = 1.$

Thus, B is the correct answer.

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Since the answer should be same regardless of the number of problems, we can assume that there were $100$ problems on the assignment. $80\%$ of $50$ is $40,$ so Chloe answered $40$ questions correctly alone. $88\%$ of $100$ is $88,$ so Chloe answered $88$ questions correctly in total. This means that Chloe answered $88 - 40 = 48$ together with Zoe.

$90\%$ of $50$ is $45,$ so Zoe answered $45$ questions correct by herself. We know that she answered $48$ questions correct with Chloe, so she answered $45 + 48 = 93$ correctly in total. This means that her overall percentage is $93\%.$

Thus, C is the correct answer.

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Starting from $A,$ there are $4$ ways to reach an $M.$ From each $M,$ there are $3$ ways to reach a $C.$ From each $C,$ there are $2$ ways to reach an $8.$ Multiplying all these possibilities, we get $4 \cdot 3 \cdot 2 = 24.$

Thus, D is the correct answer.

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The only way to split $\overline{BC}$ into two parts such that the two triangles have the same perimeter is if $\overline{CD} = 3$ and $\overline{BD} = 2.$

$\triangle ACD$ and $\triangle ABD$ have the same altitudes, so their areas are proportional to their bases. This means that the area of $\triangle ABD$ is $\dfrac{2}{5}$ the area of $\triangle ABC,$ which is $$\dfrac{2}{5} \cdot 3 \cdot \dfrac{4}{2} = \dfrac{12}{5}.$$

Thus, D is the correct answer.

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Let $n$ be the number of treasure chests and $g$ be the number of gold coins. Then $$9(n - 2) = g$$ and $$6n + 3 = g.$$ Solving this system yields $n = 7,$ so the number of gold coins is $6 \cdot 7 + 3 = 45.$

Thus, C is the correct answer.

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Since $\angle BCD$ is a right angle, we can apply the Pythagorean theorem to $\triangle BCD$ to get that $\overline{BD} = 5.$ We also get that $\angle DBA$ is right since the sides of$\triangle BDA$ form a Pythagorean triple.

Then the area of $ABCD$ is equal to $$\begin{align*} \text{area}(\triangle &BDA) - \text{area}(\triangle BCD) \\ &= \dfrac{1}{2} \cdot 12 \cdot 5 - \dfrac{1}{2} \cdot 4 \cdot 3 \\ &= 30 - 6 \\ &= 24. \end{align*}$$

Thus, B is the correct answer.

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The expression can be factored as follows:

$$\begin{align*} &98! + 99! + 100! \\ &= 98! + 99 \cdot 98! + 100 \cdot 99 \cdot 98! \\ &= 98!(1 + 99 + 100 \cdot 99) \\ &= 98!(100 + 100 \cdot 99) \\ &= 98! \cdot 100 \cdot (1 + 99) \\ &= 98! \cdot 100^2 \end{align*}$$

Each $100$ possesses two factors of $5.$ The number of factors of $5$ in $98!$ is $\lfloor 98 / 5 \rfloor + \lfloor 98 / 25 \rfloor= 19 + 3 = 22.$ This counts the number of numbers divisible by $5$ and the number of numbers divisible by $25$ to get both factors of $5.$ The total number of factors of $5$ is $2 + 2 + 22 = 26.$

Thus, D is the correct answer.

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Since the number is odd, the last digit is odd, giving $5$ possibilities. The thousands digit cannot be zero or the the number we already got, so that gives $8$ possibilities. Similarly, the hundreds digit has $8$ possibilities, and the tens digit has $7$ possibilities. This gives a total of $5 \cdot 8 \cdot 8 \cdot 7 = 2240,$ making the probability $\dfrac{2240}{9000} = \dfrac{56}{225}.$

Thus, B is the correct answer.

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Since the sum is $0,$ all the numbers cannot be positive or negative. This gives two cases: two are positive and one is negative or vice versa. Also note that if $x < 0,$ $\dfrac{x}{|x|} = -1,$ and if $x > 0,$ $\dfrac{x}{|x|} = 1.$

$\textbf{Case I:}$ two are positive and one is negative

WLOG, let $a, b > 0$ and $c < 0.$ Then $abc < 0.$ This means that $\dfrac{a}{|a|} = \dfrac{b}{|b|} = 1$ and $\dfrac{c}{|c|} = \dfrac{abc}{|abc|} = -1.$ Then $$\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|} = 0.$$

$\textbf{Case II:}$ two are negative and one is positive

WLOG, let $a, b < 0$ and $c > 0.$ Then $abc > 0.$ This means that $\dfrac{a}{|a|} = \dfrac{b}{|b|} = -1$ and $\dfrac{c}{|c|} = \dfrac{abc}{|abc|} = 1.$ Then $$\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|} = 0.$$

Either way, $\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|}$ equals $0.$

Thus, A is the correct answer.

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Let $O$ be the center of the inscribed semicircle and $D$ be the tangent point of the semicircle on $\overline{AB}.$ Then $BD = 5$ since $\overline{BD}$ and $\overline{BC}$ are tangents to the semicircle. Then $AD = 8$ and $OD = r.$ $\overline{OD}$ is perpendicular to$\overline{AB}$ so $\triangle ADO \sim \triangle BCA,$ so $\dfrac{r}{8} = \dfrac{5}{12}.$ Solving this, we get $r = \dfrac{10}{3}.$

Thus, D is the correct answer.

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Linda traveled for $60$ minutes every day. Since one mile was traveled in an integer amount of minutes each day, her mph every day must be a factor of $60.$ The factors of $60$ are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30,$ and $60.$ The only sequence of four of these numbers that differ by $5$ are $5, 10, 15,$ and $20.$ For the four days, she traveled $\dfrac{60}{5} + \dfrac{60}{10} + \dfrac{60}{15} + \dfrac{60}{20} = 25$ miles.

Thus, C is the correct answer.

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In a $60$ day period, the first child calls $20$ times, the second child calls $15$ times, and the third child calls $12$ days. $20 + 15 + 12 = 47$ overcounts, however. The first and second children call on the same day $60 / 12 = 5$ times. The first and third children call on the same day $60 / 15 = 4$ times. The second and third children call on the same day $60 / 20 = 3$ times. Subtracting these from $47$ yields $47 - 5 - 4 - 3 = 35.$

The $60^{\text{th}}$ is added in thrice and subtracted out thrice, so we need to add it back in. This means that for every $60$ days, Mrs. Sanders receives a call $36$ days, which means that she does not receive a call on $24$ days. There are $6$ $60$ day periods, and there are no calls the $361^{\text{st}}$ or $362^{\text{nd}}$ day, which results in $24 \cdot 6 + 2 = 146$ total days with no phone calls.

Thus, D is the correct answer.

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We can extend $\overline{SU}$ and $\overline{TU}$ to form the following picture.

The area of this region is the area of an equilaterial triangle with side length of $4$ minus the area of two-sixths of a circle with radius $2.$ The area for an equilateral triangle with side length $s$ is $\dfrac{s^2\sqrt{3}}{4}.$ This means that the total area is $\dfrac{4^2 \sqrt{3}}{4} - \dfrac{1}{3} \pi \cdot 2^2 = 4 \sqrt{3} - \dfrac{4}{3} \pi.$

Thus, B is the correct answer.