Question

Rosie went on a hiking trip. The first day she walked 181818 kilometers. Each day since, she walked 90\%90%90, percent of what she walked the day before. What is the total distance Rosie has traveled by the end of the 10^\text{th}10 th 10, start superscript, t, h, end superscript day? Round your final answer to the nearest kilometer.

Answer 1

Answer:

117 km

Explanation:

The 1st day, Rosie walked

$d_1 = 18 km$

The 2nd day, she walked 90% of the distance walked the previous day, so

$d_2 = 0.90d_1 = 0.90 (18)$

Similarly, on the 3rd day she walked

$d_3 = 0.90 d_2 = 0.90 (0.90)(18)=(0.90)^2 (18)$

So the distance walked at the nth day is

$d_n = (0.90)^{n-1} 18$

This is a geometrical series of the form

$d_n = d_1 r^{n-1}$

where $d_1 = 18$ and $r=0.90$. The sum of such a series is given by

$\sum d_n = d_1(\frac{1-r^n}{1-r})$

So for n = 10, we find:

$\sum d_{10}= 18(\frac{1-0.90^{10}}{1-0.90})=117 km$

Answer 2

We will see that the total distance walked at the end of the 10 days is 117km

What is the total distance Rosie walked?

We know that on the first day, she walks a distance of 18 km.

The next day, she walks 90% of that, so the distance walked on day 2 is:

d(2) = (90%/100%)*18km = 0.9*18km

On day 3, she walks the 90% of what she walked on day 2, so the distance is:

d(3) = (90%/100%)*0.9*18km = 0.9^2*18km

We already can see tha pattern here, on day x, she walks a distance:

d(x) = 18km*(0.9)^(x - 1)

This is an exponential decay.

Then the total distance walked on the first 10 days is:

D = 18km*(1 + 0.9 + 0.9^2 + 0.9^3 + ... + 0.9^9) = 117 km

If you want to learn more about exponential decays, you can read:

https://brainly.com/question/11464095