1. One year the Seattle Mariners won 116 games and lost 46. What was their winning percentage? (Hint: Divide the number of games won by the total number of games played.)

2. If the body of a 155-pound person contains 93 pounds of water, what percent of his body is water?

Calculating the Winning Percentage
The formula for winning percentage is:
[tex]\text{Winning Percentage} = \left(\frac{\text{Number of Games Won}}{\text{Total Number of Games Played}}\right) \times 100[/tex]
In this case, the Seattle Mariners won 116 games and lost 46 games. Therefore, the total number of games played is:
[tex]\text{Total Games} = 116 + 46 = 162[/tex]
Now, plug the numbers into the formula:
[tex]\text{Winning Percentage} = \left(\frac{116}{162}\right) \times 100[/tex]
Doing the division:
[tex]\frac{116}{162} \approx 0.716[/tex]
And finally, multiply by 100 to convert it into a percentage:
[tex]\text{Winning Percentage} \approx 71.6\%[/tex]
Thus, the Seattle Mariners' winning percentage that year was approximately 71.6%.
Calculating the Percent of Body Water Weight
To find out what percent of the body weight is water, use this formula:
[tex]\text{Percentage of Body as Water} = \left(\frac{\text{Water Weight}}{\text{Total Body Weight}}\right) \times 100[/tex]
In this situation, we have:

Plugging in the values:

[tex]\text{Percentage of Body as Water} = \left(\frac{93}{155}\right) \times 100[/tex]

Doing the division:

[tex]\frac{93}{155} \approx 0.6[/tex]

Multiply by 100:

[tex]\text{Percentage of Body as Water} \approx 60\%[/tex]

Therefore, approximately 60% of the person's body weight is made up of water.

1. One year the Seattle Mariners won 116 games and lost 46. What was their winning percentage? (Hint: Divide the number of games won by the total number of games played.)2. If the body of a 155-pound person contains 93 pounds of water, what percent of his body is water?