0.5t+12=t+10.50------set the equations equal to eachother, with t being the number of toppings
1.5=0.5t----subtract 0.5t from each side and 10.5
t=3---if there are 3 toppings the prices will be the same
This is hard in mathematical language, but there is only this way to present it. Let us denote by A the consentration of the substance and by A' its rate.
We have that b/A= t+c where c is a constant.
Hence A=b/(c+t)
By differentiating, we get that A'=-b/(C+t)^2
Then, we have that -(A)^2=A'/b
Hence at any point, we have that if we make the concentration 17 times, we have that there is a 17 times bigger , the rate will become bigger by 17*17, hence 289 times faster.
You left it out, but I'm thinking that there must be an 'x' next to the '20.50' in the function. I'm so sure of it that I'll assume it, as I proceed to answer the question:
C(x) = 20.50x + 2,000
Subtract 2,000 from each side: C - 2,000 = 20.50 x
Divide each side by 20.50 : x = (C - 2,000) / 20.50
When C = $625,000 . . .
x = (625,000 - 2,000) / 20.50 = 623,000 / 20.50 = 30,390.2439
<em>30,390 complete units</em> are produced, and there are 5 bucks left over,
to split up among all the loyal employees who worked with such diligence and dedication to make it happen. The company's senior management will graciously add each worker's share to his gross pay before taxes for the second month following the close of the current quarter, with a photocopied note inserted in the pay envelope, expressing management's sincere thanks to everyone, an admonition not to spend it all in one place, and a reminder that no matter how many festivals to their god they need to go out to the desert to celebrate, their tally of bricks for the next quarter shall not be diminished.
Answer:
The interval [32.6 cm, 45.8 cm]
Step-by-step explanation:
According with the <em>68–95–99.7 rule for the Normal distribution:</em> If 
is the mean of the distribution and s the standard deviation, around 68% of the data must fall in the interval
![\large [\bar x - s, \bar x +s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-%20s%2C%20%5Cbar%20x%20%2Bs%5D)
around 95% of the data must fall in the interval
![\large [\bar x -2s, \bar x +2s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-2s%2C%20%5Cbar%20x%20%2B2s%5D)
around 99.7% of the data must fall in the interval
![\large [\bar x -3s, \bar x +3s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-3s%2C%20%5Cbar%20x%20%2B3s%5D)
So, the range of lengths that covers almost all the data (99.7%) is the interval
[39.2 - 3*2.2, 39.2 + 3*2.2] = [32.6, 45.8]
<em>This means that if we measure the upper arm length of a male over 20 years old in the United States, the probability that the length is between 32.6 cm and 45.8 cm is 99.7%</em>
85 is a composite number because 17×5=85
