Hello,
Your correct answer would be A.
Hope this helps!
I believe the correct answer is B
We know form our problem that the third day she biked 20 miles, so we have the point (3,20). We also know that <span>on the eighth day she biked 35 miles, so our second point is (8,35).
To relate our two point we are going to use the slope formula: </span>
![m= \frac{x_{2}-x_{1}}{y_{2}-y_{1}}](https://tex.z-dn.net/?f=m%3D%20%5Cfrac%7Bx_%7B2%7D-x_%7B1%7D%7D%7By_%7B2%7D-y_%7B1%7D%7D%20)
We can infer form our points that
![x_{1}=3](https://tex.z-dn.net/?f=x_%7B1%7D%3D3)
,
![y_{1}=20](https://tex.z-dn.net/?f=y_%7B1%7D%3D20)
,
![x_{2}=8](https://tex.z-dn.net/?f=x_%7B2%7D%3D8)
, and
![y_{2}=35](https://tex.z-dn.net/?f=y_%7B2%7D%3D35)
. so lets replace those values in our slope formula:
![m= \frac{35-20}{8-3}](https://tex.z-dn.net/?f=m%3D%20%5Cfrac%7B35-20%7D%7B8-3%7D%20)
![m= \frac{15}{5}](https://tex.z-dn.net/?f=m%3D%20%5Cfrac%7B15%7D%7B5%7D%20)
![m=3](https://tex.z-dn.net/?f=m%3D3)
Now that we have the slope, we can use the point-slope formula <span>determine the equation of the line that best fit the set for Maggie’s data.
Point-slope formula: </span>
![y-y_{1}=m(x-x_{1})](https://tex.z-dn.net/?f=y-y_%7B1%7D%3Dm%28x-x_%7B1%7D%29)
![y-20=3(x-3)](https://tex.z-dn.net/?f=y-20%3D3%28x-3%29)
![y-20=3x-9](https://tex.z-dn.net/?f=y-20%3D3x-9)
![y=3x+11](https://tex.z-dn.net/?f=y%3D3x%2B11)
We can conclude that the equation of the line that best fit the set for Maggie’s data is
![y=3x+11](https://tex.z-dn.net/?f=y%3D3x%2B11)
.
29 and 3
13 and 19
Hope this helps!
Answer:
Step-by-step explanation:
<u>Given:</u>
- Investment P = $20000
- Time t = 7 years
- Interest rate r = 5.5% = 0.055
a. <u>compounded semiannually, n = 2</u>
b. <u>compounded quarterly, n = 4</u>
c. <u>compounded monthly, n = 12</u>
d. <u>compounded continuously</u>