Answer:
Step-by-step explanation:
Volume of the butter in the cardboard = 31.25 cubic inches
Length of the Molded Square Prism =5 Inches
Let the length of the square cross sectional area = x
Volume of the Prism
Since l=5 Inches

Next, we determine the surface area of the New Cardboard box to be made.
Length of the Cardboard Box=5 Inches
Length of the square cross sectional area = 2.5 Inches
Total Surface Area

Therefore,
of cardboard is needed,
Answer:
The probability of NOT hitting a boundary is (4/5).
Step-by-step explanation:
Let E: Be the event of hitting a boundary
now, Probability of any event E = 
Here, number of favorable outcomes = 6
So, P(E) = 
⇒Probability of hitting a six is 1/5
Now, P(E) + P(not E) = 1
So, P(not hitting a boundary ) = 1 - P(hitting a boundary)
= 1 - (1/5) = 4/5
Hence, the probability of NOT hitting a boundary is (4/5).
Step-by-step explanation:
-2(-3+2)2
(-2×-3)+(-2×2)2
(6-4)2
2(6-4)
(2×6)+(2×-4)
12-8=4
Answer:
80.66
Step-by-step explanation:


Since we have three variables,



Therefore:

Using the above:
f(4.02, 3.99, 6.97) =81+ 8(4.02 - 4) + 8 (3.99 - 4) +14(6.97-7)=80.66.
The approximation of
is 80.66000.