put a 1 under the 13, so it is 13/1, and 5 5/6, turn it into an improper fraction by multiplying 5x6=30. then 30+5, which is 35. so just keep the denominator, which would be 35/6. so you would have 13/1 - 35/6. you need the same denominator, so see what both 1 and 6 would fit into. which is 6. so 13x6 which is 78, so 78/6 - 35/6, so 78-35 is 43. so 43/6, then simplify, which is 7 1/6
Answer:
We see that the only complete time-distance pair indicates that she walked 6.4 miles in 2 hours. If she walked at a constant speed, we can conclude that girl walked 3.2 miles in 1 hour. Using this speed, we can find the remaining values in the table. The easier entry to complete is finding out how far she walked in 5 hours: 5 hours⋅3.2 miles/hour=16 miles. To find out how long it took to walk 8 hours we can either notice that 8 is half of 16, so it took half the time to walk half the distance or we can divide 8 miles by 3.2 miles per hour to find that it took 2.5 hours
Step-by-step explanation:
Answer:
Following are the solution to the given choices:
Step-by-step explanation:
Using chebyshev's theorem
:
![\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44](https://tex.z-dn.net/?f=%5Cto%20P%28%7C%28x-%5Cmu%29%7C%5Cleq%20k%20%5Csigma%29%5Cgeq%201-%20%5Cfrac%7B1%7D%7Bk%5E2%7D%5C%5C%5C%5Chere%20%5C%5C%20%5Cto%20%5Cmu%3D28.75%5C%5C%5C%5C%20%5Cto%20%5Csigma%3D4.44)
In point a)
![\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}](https://tex.z-dn.net/?f=%5Cto%20%7C%28x-%5Cmu%29%7C%3D%20%7C20.17-28.75%7C%3D8.58%20and%20%5Csigma%3D4.44%20%5C%5C%5C%5Cso%2C%20%5C%5Ck%3D%20%5Cfrac%7B%20%7C%28x-%5Cmu%29%7C%20%7D%7B%5Csigma%7D)
![= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B8.58%7D%7B4.44%7D%20%5C%5C%5C%5C%20%3D1.9%20%5C%5C%5C%5C%20%3D2)
![value= (1- \frac{1}{k^2}) \times 100 \% =75\%](https://tex.z-dn.net/?f=value%3D%20%281-%20%5Cfrac%7B1%7D%7Bk%5E2%7D%29%20%5Ctimes%20100%20%5C%25%20%3D75%5C%25)
In point b)
![\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)](https://tex.z-dn.net/?f=%5Cto%20%281-%20%5Cfrac%7B1%7D%7Bk%5E2%7D%29%20%5Ctimes%20100%20%5C%25%20%3D84%5C%25%20%5C%5C%5C%5C%5Cto%20%281-%20%5Cfrac%7B1%7D%7Bk%5E2%7D%29%3D0.84%20%5C%5C%5C%5C%5Cto%20%5Cfrac%7B1%7D%7Bk%5E2%7D%20%3D0.16%20%5C%5C%5C%5C%5Cto%20%5Cfrac%7B1%7D%7Bk%7D%3D0.4%5C%5C%5C%5C%20%5Cto%20k%3D2.5%20%5C%5C%5C%5C%5Cto%20%7C%28x-%5Cmu%29%7C%20%5Cleq%20k%20%5Csigma%20%20%5C%5C%5C%5C%3D%20%7C%28x-%5Cmu%29%7C%5Cleq%202.5%20%5Ctimes%204.44%20%20%5C%5C%5C%5C%20%20%3D%20%7C%28x-%5Cmu%29%7C%5Cleq%201.11%20%5C%5C%5C%5C%20%3D%20%20%2828.75%5Cpm%2011.1%29%20%5C%5C%5C%5C%5Cto%20%5Ctext%7Bfares%20lies%20between%7D%2817.65%2C39.85%29)
In point c)
![\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\](https://tex.z-dn.net/?f=%5Cto%2099.7%20%5C%25%20%5C%5C%20lie%20%5C%20between%3D28.75%20%5Cpm%20z%28.03%29%5Ctimes%20%5Csigma%20%5C%5C%5C%5C%20%3D28.75%5Cpm%202.97%20%5Ctimes%204.44%5C%5C%5C%5C%3D%2828.75%5Cpm%2013.1%29%5C%5C%5C%5C%3D%2815.65%2C41.85%29%5C%5C)
In point d)
using standard normal variate
![x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17](https://tex.z-dn.net/?f=x%3D20.17%5C%5C%5C%5C%20%20z%3D-2%20%5C%5C%5C%5Cx%3D37.13%5C%5C%5C%5C%20z%3D2%5C%5C%5C%5C%5Cto%20P%2820.17%3Cx%3C37.13%29%3DP%28-2%3Cz%3C2%29%3D0.95%20%5C%5C%5C%5C)
Joseph Pulitzer
Rough Riders
The Philippines