Answer:
f(t) = 5×0.87^t
Step-by-step explanation:
The general form for an exponential function described in this fashion is ...
... f(t) = (starting value) × (1 + (percent change))^t
Here, the "percent change" is -13%, or -0.13.
Then the value (1 + percent change) is (1 + (-0.13)) = 0.87. Putting this and the starting value into the form above, we have ...
... f(t) = 5 × 0.87^t
Step-by-step explanation:
Since it remains only 1 sweet, we can subtract it from the total and get the amount of sweets distributed (=1024).
As all the sweets are distributed equally, we must divide the number of distributed sweets by all its dividers (excluding 1024 and 1, we'll see later why):
1) 512 => 2 partecipants
2) 256 => 4 partecipants
3) 128 => 8 partecipants
4) 64 => 16 partecipants
5) 32 => 32 partecipants
6) 16 => 64 partecipants
7) 8 => 128 partecipants
9) 4 => 256 partecipants
10) 2 => 512 partecipants
The number on the left represents the number of sweets given to the partecipants, and on the right we have the number of the partecipants. Note that all the numbers on the left are dividers of 1024.
Why excluding 1 and 1024? Because the problem tells us that there remains 1 sweet. If there was 1 sweet for every partecipant, the number of partecipants would be 1025, but that's not possible as there remains 1 sweet. If it was 1024, it wouldn't work as well because the sweets are 1025 and if 1 is not distributed it goes again against the problem that says all sweets are equally distributed.
Answer:
20(pi)
Step-by-step explanation:
• equation for the circumference of a circle= 2(pi)r
• You are given the diameter ( 20ft) but you need the radius (r) for the equation.
• divide 20 by 2 to get the radius since it is half of the diameter.
• 20/2=10 so c= 2(pi)(10)
• 2(10)=20
• plug in pi and your final answer is 20(pi)
Answer:
The correct answer will be X = 4
Step-by-step explanation:
Using Pythagoras
AB^2 + BC^2 = AC^2
2^2 + (2√3)^2 = AC^2
4 + (4*3) = AC^2
4+12 = AC ^ 2
AC ^2 = 16
AC = √16 = 4
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
