I believe the answer is B
let's firstly change the 1.2 to a fraction
![1.\underline{2}\implies \cfrac{12}{1\underline{0}}\implies \cfrac{6}{5} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{\frac{6}{5}}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{\frac{6}{5}}-\stackrel{y1}{6}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{10}}}\implies \cfrac{~~ \frac{6-30}{5}~~}{-6}\implies \cfrac{~~ \frac{-24}{5}~~}{-6}\implies \cfrac{~~ -\frac{24}{5}~~}{-\frac{6}{1}}](https://tex.z-dn.net/?f=1.%5Cunderline%7B2%7D%5Cimplies%20%5Ccfrac%7B12%7D%7B1%5Cunderline%7B0%7D%7D%5Cimplies%20%5Ccfrac%7B6%7D%7B5%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B10%7D~%2C~%5Cstackrel%7By_1%7D%7B6%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B%5Cfrac%7B6%7D%7B5%7D%7D%29%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B%5Cfrac%7B6%7D%7B5%7D%7D-%5Cstackrel%7By1%7D%7B6%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B4%7D-%5Cunderset%7Bx_1%7D%7B10%7D%7D%7D%5Cimplies%20%5Ccfrac%7B~~%20%5Cfrac%7B6-30%7D%7B5%7D~~%7D%7B-6%7D%5Cimplies%20%5Ccfrac%7B~~%20%5Cfrac%7B-24%7D%7B5%7D~~%7D%7B-6%7D%5Cimplies%20%5Ccfrac%7B~~%20-%5Cfrac%7B24%7D%7B5%7D~~%7D%7B-%5Cfrac%7B6%7D%7B1%7D%7D)
![-\cfrac{\stackrel{4}{~~\begin{matrix} 24 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{5}\cdot -\cfrac{1}{\underset{1}{~~\begin{matrix} 6 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\implies \boxed{\cfrac{4}{5}}](https://tex.z-dn.net/?f=-%5Ccfrac%7B%5Cstackrel%7B4%7D%7B~~%5Cbegin%7Bmatrix%7D%2024%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%7D%7B5%7D%5Ccdot%20-%5Ccfrac%7B1%7D%7B%5Cunderset%7B1%7D%7B~~%5Cbegin%7Bmatrix%7D%206%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%7D%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B4%7D%7B5%7D%7D)
Answer:
1/3
Step-by-step explanation:
Range of data set: 12 - 6 = 6
Given that there is a uniform distribution:
P(6<x<7) = 1/6
P(7<x<8) = 1/6
P(8<x<9) = 1/6
P(9<x<10) = 1/6
P(10<x<11) = 1/6
P(11<x<12) = 1/6
where x is weight loss
Probability that weight loss is more than 10 pounds:
P(10<x<11) + P(11<x<12) = 1/6 + 1/6 = 1/3
If you do 5 divide by 4 then you can find what 1$ would be worth so ie.
5 divide by 4 is 1.25 which means that for 1$ you could buy 1.25 ounces
<span>
(1/6)/ 1.5 =
(1/6)/(15/10)=
1/6*10/15=10/90=1/9 width</span>