Pre-Algebra

1st Edition

ISBN: 9780618800766

Author: Ron Larson, Lee Stiff, Laurie Boswell, Timothy Kanold

Publisher: Houghton Mifflin-High School

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Question

Chapter 1.1, Problem 52E

To determine

To write the variable expression for the total number of points.

Expert Solution

Answer to Problem 52E

The variable expression for total number of points is obtained as:

3⋅f+1⋅p

Explanation of Solution

Given information:

Each field goal (FG) is equal to 3 points and each kicked point after touchdown (PAT) is equal to 1 point.

Calculation:

Let, p and f be the number of PAT and number of FG that a kicker scored respectively.

Since, each field goal (FG) is equal to 3 points and each kicked point after touchdown (PAT) is equal to 1 point.

Hence,

The variable expression for total number of points is obtained as:

3⋅f+1⋅p

To determine

To-determine the total number of points for each kicker.

Expert Solution

Answer to Problem 52E

Total number of FGs and PATs scored by George Blanda is 1948 .

Total number of FGs and PATs scored by Nick Lowery is 1711 .

Total number of FGs and PATs scored by Norm Johnson is 1736 .

Explanation of Solution

Given information:

Each field goal (FG) is equal to 3 points and each kicked point after touchdown (PAT) is equal to 1 point.

From the table provided in the question,

Number of FGs and PATs scored by George Blanda is 335 and 943 respectively.

Number of FGs and PATs scored by Nick Lowery is 383 and 562 respectively.

Number of FGs and PATs scored by Norm Johnson is 366 and 638 respectively.

Calculation:

Since, each field goal (FG) is equal to 3 points and each kicked point after touchdown (PAT) is equal to 1 point.

Hence,

Number of FGs and PATs scored by George Blanda is 335 and 943 respectively.

So,

Total number of FGs and PATs scored by George Blanda is obtained as:

3⋅335+1⋅943=1005+943 [Multiply]=1948 [Add]

Number of FGs and PATs scored by Nick Lowery is 383 and 562 respectively.

So,

Total number of FGs and PATs scored by Nick Lowery is obtained as:

3⋅383+1⋅562=1149+562 [Multiply]=1711 [Add]

Number of FGs and PATs scored by Norm Johnson is 366 and 638 respectively.

So,

Total number of FGs and PATs scored by Norm Johnsonis obtained as:

3⋅366+1⋅638=1098+638 [Multiply]=1736 [Add]

To determine

List the players from least total number of points to greatest.

Expert Solution

Answer to Problem 52E

The player can be listed as:

Player	Total number of points
George Blanda	1948
Nick Lowery	1711
Norm Johnson	1736

Explanation of Solution

Given information:

Each field goal (FG) is equal to 3 points and each kicked point after touchdown (PAT) is equal to 1 point.

From the table provided in the question,

Number of FGs and PATs scored by George Blanda is 335 and 943 respectively.

Number of FGs and PATs scored by Nick Lowery is 383 and 562 respectively.

Number of FGs and PATs scored by Norm Johnson is 366 and 638 respectively.

Calculation:

Since, each field goal (FG) is equal to 3 points and each kicked point after touchdown (PAT) is equal to 1 point.

Hence,

Number of FGs and PATs scored by George Blanda is 335 and 943 respectively.

So,

Total number of FGs and PATs scored by George Blanda is obtained as:

3⋅335+1⋅943=1005+943 [Multiply]=1948 [Add]

Number of FGs and PATs scored by Nick Lowery is 383 and 562 respectively.

So,

Total number of FGs and PATs scored by Nick Lowery is obtained as:

3⋅383+1⋅562=1149+562 [Multiply]=1711 [Add]

Number of FGs and PATs scored by Norm Johnson is 366 and 638 respectively.

So,

Total number of FGs and PATs scored by Norm Johnson is obtained as:

3⋅366+1⋅638=1098+638 [Multiply]=1736 [Add]

Hence,

The player can be listed as:

Player	Total number of points
George Blanda	1948
Nick Lowery	1711
Norm Johnson	1736

Chapter 1.1, Problem 51E

Chapter 1.1, Problem 53E

Chapter 1 Solutions

Pre-Algebra

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Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - Prob. 51E Ch. 1.1 - Prob. 52E Ch. 1.1 - Prob. 53E Ch. 1.1 - Prob. 54E Ch. 1.1 - Prob. 55E Ch. 1.1 - Prob. 56E Ch. 1.1 - 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Intro Stats, Books a la Carte Edition (5th Edition)

Suppose that A and B are mutually exclusive events for which P(A) = .3 and P(B) = .5. What is the probability t...

A First Course in Probability (10th Edition)

Fill in each blank so that the resulting statement is true. If n is a counting number, bn, read ______, indicat...

College Algebra (7th Edition)

the given equation

Pre-Algebra Student Edition

3. Voluntary Response Sample What is a voluntary response sample, and why is such a sample generally not suitab...

Elementary Statistics

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