Answer:
I have no idea. : /
Step-by-step explanation:
Vbox-vspheres
vbox=, I assume we are dealing with l=w=h
so
v=lwh=12^3=1728
vsphere=(4/3)pir^3
r=3
vsphere=(4/3)pi3^3=4pi9=36pi
8 of them so
8 times 36pi=288pi or about 904.7786842338604526772412943845 cubic inches
vbox-sphere=1728-904.7786842338604526772412943845=823.2213157661395473227587056155
space filled by packing beads is about 823.22 cubic inches
beads percent is 823.2213157661395473227587056155/1728 times 100=47.64% filled by beads
Answer:
Below
I hope its not too complicated

Step-by-step explanation:




Answer:
164
The 982 part seems wrong. Are you sure it isn't 981?
around 436 if I use 981 instead. 982 would give a decimal and that would be kinda weird for a part human.
Step-by-step explanation:
You can divide the 205 by 5 to simplify the ratio down. 205/5 = 41. Then you can multiply by 4 to find then number of boys.
981
So you can add the 5 and 4 together to 9 and then divide 981 by 9 to get 109. Then you can multiply by 4 to get 436.
Hope this helps!
Answer:
Intersept is (4.55,0)
Step-by-step explanation: