Answer:
Class 7A collected 4.8 ounces of pure gold.
Step-by-step explanation:
<u>Key skills required are: Percentages, Multiplication</u>
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- We only need the information about Class 7A. They collected 40 oz that contains 12% gold.
- In other words, this means that 12% of that 40 oz gold sand is pure gold.
Here we have to do 12% x 40 to find the number of oz of pure gold. We first have to convert 12% into a decimal. Divide it by a 100 (or move the decimal point 2 places to the left) and you will get 0.12.
<em>Do 0.12 x 40 and you will get 4.8</em>
Therefore, there are 4.8 oz of pure gold in Class 7A's gold sand
Answer:
Step-by-step explanation:
It is the last one. A closed circle means less than or equal to, so that crosses out the top two. The third says five is greater and we know that is not true because the line is decreasing in value. It must be the fourth.
Answer:
Step-by-step explanation:
For a first order decay, fraction remaining = 0.5n where n = number of half lives elapsed.
fraction remaining = 55% = 0.55
0.55 = 0.5n
log 0.55 = n log 0.5
-0.2596 = -0.301 n
n = 0.8625 = # of half lives elapsed
0.8625 half lives x 2.5 billion years/half live = 2.16 billion years have elapsed = age of the rock
b) 0.15 = 0.5n
log 0.15 = n log 0.5
-0.824 = -0.301 n
n = 2.74 half lives
2.5 billion years/half life x 2.74 half lives = 6.85 billion years = age of rock
The maximized value of the function is (c) 119/2
<h3>Maximization problem</h3>
Maximization problems are used to determine the optimal solution of a linear programming model
<h3>Objective function</h3>
The objective function is given as:
<h3>Constraints</h3>
The constraints are given as:
<h3>Graph</h3>
See attachment for the graph of the constraints
From the graph, the optimal solution is: (2.83, 2.83)
So, the maximized value is:
Approximate
Rewrite as a fraction
Hence, the maximized value of the function is (c) 119/2
Read more about maximization problem at:
brainly.com/question/16826001
