Which of the following quadratic regression equations best fits the data shown below?
1 answer:
Answer:
Step-by-step explanation:
In order to do this, you have to know how to use your calculator's regression equation function.
First enter in the data. Hit "stat" then 1:Edit and enter all the x values into L1. After each value, hit enter. When you're done with the x list, arrow over to L2 and enter in all the y-values. If there are already values there you need to clear, arrow up to highlight L1, hit "clear", then "enter" and the values will disappear. Do that for both lists if you need to.
After the data is listed in L1 and L2, hit "stat" again and arrow over to "Calc". Hit 5:QuadReg. If you have a TI 83, your equation will be there for you. If you have a TI 84, you'll need to arrow down to "calculate" to get the equation. Regardless, the equation is
, the last choice in your options.
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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.
<h2>Answer:</h2>
In the 5th year
<h2>Step-by-step explanation:</h2>For the first year, the salary is 1.2million = 1,200,000
For the second year, the salary is 1.2million + 75000 = 1,200,000 + 75,000 = 1,275,000
.
.
.
For the last year, the salary is 1.5million = 1,500,000
This gives the following sequence...
1,200,000 1,275,000 . . . 1,500,000
This follows an arithmetic progression with an increment of 75,000.
<em>Remember that,</em>
The last term, L, of an arithmetic progression is given by;
L = a + (n - 1)d ---------------(i)
<em>Where;</em>
a = first term of the sequence
n = number of terms in the sequence (which is the number of years)
d = the common difference or increment of the sequence
<em>From the given sequence,</em>
a = 1,200,000 [which is the first salary]
d = 75,000 [which is the increment in salary]
L = 1,500,000 [which is the maximum salary]
<em>Substitute these values into equation (i) as follows;</em>
1,500,000 = 1,200,00 + (n - 1) 75,000
1,500,000 - 1,200,000 = 75,000(n-1)
300,000 = 75,000(n - 1)
4 = n - 1
n = 5
Therefore, in the 5th year the maximum salary will be reached.