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:
24ab-8ac
Step-by-step explanation:
Assuming you want to simplify it.
Original equation: 8a(3b+6c-7c)
Apply 8a to each variable in the paranthesis: (8a)(3b)+(8a)(6c)+(8a)(-7c)
After multiplication: 24ab+48ac-56ac
Combine like terms: 24ab-8ac
Answer:
-2, -13
0, -3
2, 7
4, 17
Step-by-step explanation:
To fill out the table, substitute the table x-values into the equation and the y-values will fill the table.
y = 5x - 3
y = 5(-2) - 3
y = -10 - 3
y = -13
This means the table value corresponding to -2 should be -13.
y = 5x - 3
y = 5(0) - 3
y = 0 - 3
y = -3
This means the table value corresponding to 0 should be -3.
y = 5x - 3
y = 5(2) - 3
y = 10 - 3
y = 7
This means the table value corresponding to 2 should be 7.
y = 5x - 3
y = 5(4) - 3
y = 20 - 3
y = 17
This means the table value corresponding to 4 should be 17.
Answer:

Step-by-step explanation:
Given: There are 2 classes of 25 students.
13 play basketball
11 play baseball.
4 play neither of sports.
Lets assume basketball as "a" and baseball as "b".
We know, probablity dependent formula; P(a∪b)= P(a)+P(b)-p(a∩b)
As given total number of student is 25
Now, subtituting the values in the formula.
⇒P(a∪b)= 
taking LCD as 25 to solve.
⇒P(a∪b)= 
∴ P(a∪b)= 
Hence, the probability that a student chosen randomly from the class plays both basketball and baseball is
.