Drag each tile to the correct location on the table. Each tile can be used more than once, but not all tiles will be used.
1 answer:
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Answer:
Carbon-14 loses around 10% ( 0.1 in decimal form) of its mass, after one millennium.
Then if we start with a mass A of Carbon-14, after one millennium we will have a mass equal to:
A - A*0.1 = A*(0.9)
After another millennium, we will have a mass equal to:
A*(0.9) - A*(0.9)*0.1 = A*(0.9)^2
And so on, this is an exponential decay.
We already can see the pattern here, after x millenniums, the mass of carbon-14 will be:
M(x) = A*(0.9)^x
Now, in this problem we have 600 grams of carbon-14, then the equation for the mass will be:
y = M(x) = 600g*(0.9)^x
And the graph of this equation is shown below.
Answer:
Now we can find the limits in order to determine outliers like this:
So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier
b. 3
Step-by-step explanation:
For this case we have the following summary:
represent the minimum value
represent the first quartile
represent the median
represent the third quartil
represent the maximum
If we use the 1.5 IQR we need to find first the interquartile range defined as:
Now we can find the limits in order to determine outliers like this:
So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier
b. 3