1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
VMariaS [17]
2 years ago
11

2 Doctors is _______________% of 25 Doctors what is it.

Mathematics
2 answers:
melisa1 [442]2 years ago
8 0

Answer:

8%

Step-by-step explanation:

if you divide 2 by 25, you would get 0.08, which is equal to 8%

Valentin [98]2 years ago
6 0

Answer:

8%

Step-by-step explanation:

You might be interested in
PLEASE I WILL GIVE BRAINLIEST AND I WIL REPORT ANSWERS THAT HAVE NOTHING RI DO WITH MATH THANKS ​
Rudiy27

Answer:

26.1533936612

Step-by-step explanation:

bc

h= 6 square root 3

24^2+6 square root 3^2=684

square root 684=26.1533936612

7 0
2 years ago
Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea
Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

d) 0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}

And if we simplify we got this:

p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we simplify we got:

p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And if we set equal to 0 we got:

0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

7 0
2 years ago
The ratio of the number of male lions to female lions in the animal reserve is 21:20. If there are 123 lions in the animal reser
Vanyuwa [196]
21 + 20 = 43

Total is 123, so divide by 43.

123/43 = 2.86 rounded to 3.

(females) 20 x 3 = 60 female lions 
7 0
2 years ago
The total number of running yards in a football game was less than 100. The inequality x < 100 represents the situation.
Aleks [24]

Answer:

sorry I'm late but it's b I'm on unit test on edge

4 0
2 years ago
Which equation could represent "the area of a square as a function of a side"?
Ratling [72]
Let side = x
area = x^2
6 0
2 years ago
Other questions:
  • You are scheduled to receive $16,500 in three years. when you receive it, you will invest it for nine more years at 9.5 percent
    6·1 answer
  • What is 4/5 %of 500
    7·1 answer
  • Evaluate the algebraic expression h + h when b = 3, d = 12, h = 2 and r = 7.
    15·2 answers
  • Kendra is buying bottled water for a class trip she has 16 bottles left over from last trip she buys bottles by the case to get
    10·1 answer
  • Help me plz with this
    10·1 answer
  • In circle Y, what is m∠1? 6° 25° 31° 37°
    15·1 answer
  • The approximate value of 5,700,000 x 200 is
    10·1 answer
  • Simplify 1 3/4 : 2 1/3​
    13·2 answers
  • Evaluate find the degree of a polynomial (y³-2)(y²+11)​
    13·1 answer
  • N − 9.02 = 3.85<br> what is N?
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!