body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

McGraw Hill Glencoe Algebra 1, 2012

MH

McGraw Hill Glencoe Algebra 1, 2012 View details

Preparing for Standardized Tests

Continue to next subchapter

Exercise 2 Page 819

a We are given a table that illustrates the number of coins in a piggy bank.

Coin	Number
Penny	$16$
Nickel	$18$
Dime	$20$
Quarter	$10$

We are asked to calculate the probability of a randomly selected coin being a dime. This probability can be found by the ratio of the number of dimes to the total number of coins in the piggy bank.

P (dime) = \frac{Number of dimes}{Total number of coins}

Let's substitute the values into this ratio and calculate it!

P (dime) = \frac{Number of dimes}{Total number of coins}

SubstituteValues

Substitute values

P (dime) = \frac{2 0}{1 6 + 1 8 + 2 0 + 1 0}

Add terms

P (dime) = \frac{2 0}{6 4}

DivFracByFracSameDenom

$\frac{a / 4}{b / 4} = \frac{a}{b}$

P (dime) = \frac{5}{1 6}

The probability that a randomly selected coin will be a dime is

\frac{5}{1 6} .

b We will now find the probability that a randomly selected coin will be a nickel or a quarter. Since it is not possible for a coin to be nickel and a quarter at the same time, these events are mutually exclusive. Therefore, the probability of selecting a nickel or a quarter is the sum of their individual probabilities.

P (nickel or quarter) = P (nickel) + P (quarter)

Let's first calculate the probability of a randomly selected coin being a nickel

P (nickel)

by using the same method as in Part A.

P (nickel) = \frac{Number of nickel}{Total number of coins}

SubstituteValues

Substitute values

P (nickel) = \frac{1 8}{1 6 + 1 8 + 2 0 + 1 0}

Add terms

P (nickel) = \frac{1 8}{6 4}

Let's now calculate the probability of a randomly selected coin being a quarter,

P (quarter) .

P (quarter) = \frac{Number of quarter}{Total number of coins}

SubstituteValues

Substitute values

P (quarter) = \frac{1 0}{1 6 + 1 8 + 2 0 + 1 0}

Add terms

P (quarter) = \frac{1 0}{6 4}

Finally, we will substitute the ratios

P (nickel) = \frac{1 8}{6 4}

and

P (quarter) = \frac{1 0}{6 4}

into the equation and solve it for

P (nickel or quarter) .

P (nickel or quarter) = P (nickel) + P (quarter)

$P (nickel) = \frac{1 8}{6 4}$ , $P (quarter) = \frac{1 0}{6 4}$

P (nickel or quarter) = \frac{1 8}{6 4} + \frac{1 0}{6 4}

Add fractions

P (nickel or quarter) = \frac{2 8}{6 4}

DivFracByFracSameDenom

$\frac{a / 4}{b / 4} = \frac{a}{b}$

P (nickel or quarter) = \frac{7}{1 6}

The probability that a randomly selected coin will be a nickel or a quarter is

\frac{7}{1 6} .

Subchapter links

Exercises p.819