The heat gained by the ice melted the ice and raised the temperature of the melted ice from its initial temperature to the final
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The answers to the two questions are:
1. The percent abundance in the container which has <u>100 navy</u>, <u>27 pinto</u>, and <u>173 black-eyed peas beans</u> is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed <em>peas </em>beans, respectively.
2. The <em>weighted </em>average score for the scores of 85, 75, 96 obtained from the evaluations of exams (20%), labs (75%), and homework (96%) is 84.1.
1. The percent abundance by type of bean is given by:
(1)
Where:
n: is the number of each type of beans
: is the <em>total number</em> of <em>beans </em>
The <em>total number </em>of <em>beans</em> can be calculated by adding the number of all the types of beans:
(2)
Where:
: is the number of navy beans = 100
: is the number of pinto beans = 27
: is the number of black-eyed <em>peas </em>beans = 173
Hence, the total number of beans is (eq 2):
Now, the <em>percent abundance</em> by type of bean is (eq 1):
- Navy beans
- Pinto beans
- Black-eyed peas beans
Hence, the percent abundance by type of bean is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.
2. The average score (S) can be calculated as follows:
(3)
Where:
e: is the score for exams = 85
l: is the score for lab reports = 75
h: is the score for homework = 96
: is the <em>percent</em> for<em> exams</em> = 70.0%
: is the <em>percent </em>for <em>lab reports</em> = 20.0%
: is the <em>percent </em>for <em>homework </em>= 10.0%
Then, the <u>average score</u> is:
We can see that if the score for each evaluation is 100, after multiplying every evaluation for its respective percent, the final average score would be 100.
Therefore, the <em>weighted </em>average score will be 84.1.
Find more about percents here:
I hope it helps you!