the large bag is 5/6 lb, the smaller bags will be 1/3 lb, so it should be 5/6 ÷ 1/3
![\bf \cfrac{5}{6}\div \cfrac{1}{3}\implies \cfrac{5}{\underset{2}{~~\begin{matrix} 6 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\cdot \cfrac{~~\begin{matrix} 3 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{1}\implies \cfrac{5}{2}\implies 2\frac{1}{2}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B5%7D%7B6%7D%5Cdiv%20%5Ccfrac%7B1%7D%7B3%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B%5Cunderset%7B2%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%5Ccdot%20%5Ccfrac%7B~~%5Cbegin%7Bmatrix%7D%203%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%7B1%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B2%7D%5Cimplies%202%5Cfrac%7B1%7D%7B2%7D)
Answer:
Step-by-step explanation:
From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .
So; as each place is fitted with either 0 or 1

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0
Now;
if a 0 bit and a 1 bit are equally likely
The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:
; 

Answer: x+y=
Step-by-step explanation:
you would just write it out step by step explaining wht u need to do to solve the question
Answer:
Ok what's your question? cna you include the graph?
Answer:
-0.5
Step-by-step explanation: