Miriam took a survey, asking each family in her neighborhood how many pets they have. She made a graph showing the results: [asy] size(5.5cm); draw((0,30)--(0,0)--(41,0)); for(int i=1; i<7; i+=1) {draw((0.6,5*i)--(-0.6,5*i)); draw((0.6,5*i)--(41,5*i),black+0.5+dashed); }; path a=(3,0)--(3,30)--(9,30)--(9,0)--cycle; path b=(9,0)--(9,15)--(15,15)--(15,0)--cycle; path c=(15,0)--(15,15)--(21,15)--(21,0)--cycle; path d=(21,0)--(21,30)--(27,30)--(27,0)--cycle; path e=(27,0)--(27,5)--(33,5)--(33,0)--cycle; path f=(33,0)--(33,5)--(39,5)--(39,0)--cycle; fill(a,gray); fill(b,gray); fill(c,gray); fill(d,gray); fill(e,gray); fill(f,gray); draw(a); draw(b); draw(c); draw(d); draw(e); draw(f); label(rotate(-45)*scale(0.8)*"0 pets",(5,-1),SE); label(rotate(-45)*scale(0.8)*"1 pet",(11,-1),SE); label(rotate(-45)*scale(0.8)*"2 pets",(17,-1),SE); label(rotate(-45)*scale(0.8)*"3 pets",(23,-1),SE); label(rotate(-45)*scale(0.8)*"4 pets",(29,-1),SE); label(rotate(-45)*scale(0.8)*"5 pets",(35,-1),SE); label(
+514 The chart:
The bar chart tells us that
6 families each have 0 pets,
3 families each have 1 pet,
3 families each have 2 pets,
6 families each have 3 pets,
1 family has 4 pets, and
1 family has 5 pets.
The total number of pets in all these families is \((6\cdot 0)+(3\cdot 1)+(3\cdot 2)+(6\cdot 3)+(1\cdot 4)+(1\cdot 5)=36\). The number of families is \(6+3+3+6+1+1=20\).Therefore, the average number of pets per family is \(\frac{36}{20} = \boxed{1.8}.\)
