Karl has eaten 1 part out of a possible 3 parts of the loaf of bread.  To find out how big each part is, we divide 24 by the number of parts (3).

\(\displaystyle \dpi{100} 24\div 3=8\)

So each part would be 8 slices of bread.  Karl ate 1 part, so he has eaten 8 slices of bread.

To find the part from the whole, you will need to multiply the whole (23) by the percent. In order to do this, first change the percent to a decimal (62% becomes 0.62).

\(\displaystyle 23 \times 0.62=14.26\)

The question askes you to round to the nearest tenths place: 14.26 becomes 14.3.

Therefore, 62% of 23 is 14.3.

There are \(\displaystyle \small 18\) students in Ms. Jones' class.  Since the probability of her calling on a male student is \(\displaystyle \tiny \frac{2}{3}\), we must first find \(\displaystyle \tiny \frac{2}{3}\) of \(\displaystyle \small 18\).

\(\displaystyle \frac{2}{3} \cdot \frac{18}{1}= \frac{36}{3}\)

Now we must reduce \(\displaystyle \tiny \frac{36}{3}\) by dividing both the top and bottom by \(\displaystyle 3\).

\(\displaystyle \frac{(36/3)}{(3/3)}= \frac{12}{1}\)

\(\displaystyle \small \frac{12}{1}\) is the same as \(\displaystyle \small 12\), so there are \(\displaystyle \small 12\) male students in Ms. Jones' class. In order to find the number of female students in Ms. Jones' class, we must subtract \(\displaystyle \small 12\) from the total number of students in the class, which is \(\displaystyle \small 18\).

\(\displaystyle \small 18-12=6\)

There are \(\displaystyle \small 6\) female students in Ms. Jones' class.

First, we need to find \(\displaystyle 50\%\) of \(\displaystyle 220\). Multiply \(\displaystyle .5\times220\) to get \(\displaystyle 110\).

Then, we need to find \(\displaystyle 10\%\) of \(\displaystyle 110\). Multiply \(\displaystyle .1\times110\) to get \(\displaystyle 11\).

All of the answer choices are factors of seventy, but only one is not prime.

A prime number is a number greater than one that has only one and itself as divisors.

Thirty-five is not a prime number, because in addition to being divisible by one and itself, it is also divisible by five and seven. Therefore, the answer is thirty-five.

To solve this problem, begin by finding the total number of students that voted for tennis, basketball, or swimming.

\(\displaystyle 8+4+2=14\)

We are told there are twenty students in the class, and now we know that fourteen of those twenty voted for the three sports mentioned. The remaining students voted for other sports.

\(\displaystyle 20-14=6\)

Six out of the twenty students, or \(\displaystyle \frac{6}{20}\) of the class, voted for other sports. 

When we reduce \(\displaystyle \frac{6}{20}\) to simplest form (divide the numerator and denominator by two), we get the fraction \(\displaystyle \frac{3}{10}\)   .

Therefore, \(\displaystyle \frac{3}{10}\) of the students voted for other sports.

\(\displaystyle \frac{1}{8} \%\) of \(\displaystyle \frac{1}{25 }\)  can be rewritten as

\(\displaystyle \frac{\frac{1}{8}}{100} \times \frac{1}{25}\)

\(\displaystyle =\frac{1}{8} \times \frac{1}{100}\times \frac{1}{25}\)

\(\displaystyle =\frac{1}{8 \times 100 \times25} = \frac{1}{20,000}\)

 \(\displaystyle \frac{1}{10} \%\) of \(\displaystyle \frac{1}{10}\) can be rewritten as 

\(\displaystyle \frac{\frac{1}{10}}{100} \times \frac{1}{10}\)

\(\displaystyle = \frac{1}{10} \times \frac{1}{100} \times \frac{1}{10} = \frac{1}{10 \times 100 \times 10} = \frac{1}{10,000}\)

 \(\displaystyle 1 \frac{1}{3} \%\) of \(\displaystyle \frac{1}{333}\) can be rewritten as 

\(\displaystyle \frac{1 \frac{1}{3}}{100} \times \frac{1}{333} = \frac{4}{3} \times \frac{1}{100} \times \frac{1}{333}= \frac{4 \times 1 \times 1}{3 \times 100 \times 333} = \frac{4}{99,900}\)

25% of 20 is equal to \(\displaystyle 20\cdot.25=5\). Therefore, if the eyeshadow cost $20, it would be discounted by $5 (the equivalent of 25% off). 

Given that \(\displaystyle 20-5=15\), the correct answer is \(\displaystyle 20\).