To solve this, we use the formula for the volume of the frustum of a cone which is expressed as:
<span>V= (H/6)[WL + (W+a)(L+b) + ab]
</span>V = 3/6 [(6 ft. × 8 ft.) + (6 + 14)(18 + 8) + (<span>14 ft. × 18 ft.)]
</span>V = 409.98 ft^3 <-----OPTION C
$38.05 or more because you need a minimum of $38.10
Answer:







Step-by-step explanation:
Rational numbers:
-are all numbers you can write as a quotient of integers
, 
-include terminating decimals. For example, 
-include repeating decimals. For example, 
Irrational numbers:
-have decimal representations that neither terminate nor repeat. For example, 
-cannot be written as quotients of integers
Answer:Wish i could help, but I'm just as confused.
Step-by-step explanation:
Answer:
a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55
b)0.6004
c)19.607
Step-by-step explanation:
Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2
X ~ Poisson(A) where 
a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55
So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55
b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment
Let X denotes the number of grams to be eaten before another fragment is detected.

c)The expected number of grams to be eaten before encountering the first fragments :
s