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
Olenka [21]
4 years ago
15

Among the 1013 respondents, 407 said that pesticides are "not at all annoying". What percentage of respondents said pesticides a

re "not at all annoying"?
Mathematics
1 answer:
Zolol [24]4 years ago
5 0
You take the number of respondents who said pesticides are “not at all annoying” and divide it by the number of total respondents. So you would divide 407 by 1013 (407/1013) then multiply what you get by 100. You get 40%
You might be interested in
Find the common ratio of the geometric sequence 1.5, 4.5, 13.5, 40.5, …
Viktor [21]

Answer:

The common ratio is 3.

Step-by-step explanation:

Take 4.5 and divide it by 1.5 and you find that the common ratio is 3. To ensure this is the answer you can multiple 4.5 by 3 and you'll get 13.5

6 0
3 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
3 years ago
For a sphere with a radius of 8, what is the surface area
stepan [7]

Answer:

64

Step-by-step explanation:

4 0
4 years ago
Which equation can be used to determine the distance between the origin and (-2, -4)?
sveticcg [70]

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad \underset{origin}{(\stackrel{x_2}{0}~,~\stackrel{y_2}{0})}\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[0 - (-2)]^2 + [0 - (-4)]^2}

7 0
3 years ago
I need these answers, please helppp
aliina [53]

Answer:

Prob of d>4: 0.333

Step-by-step explanation:

180*.333=60

4 0
3 years ago
Other questions:
  • Bax invested a total of $2000 in two simple interest accounts. Account A earns 3% interest and Account B earns 5% interest. Bax
    15·1 answer
  • Brenlyn's boss told her that once she has worked with the company for 10 years, her salary will be increased to 3 times what she
    13·2 answers
  • How many triangles can be constructed with angles measuring 50º, 90º, and 40º?
    15·2 answers
  • Solve the equation
    13·2 answers
  • Find the equation of the line: With an x intercept of 4 and ay intercept of −1.5.
    5·2 answers
  • Calculate the standard deviation of the data set below. (7, 9, 10, 11, 13) The standard deviation is 4. The standard deviation i
    12·2 answers
  • O-O) ................. (>_<)
    9·1 answer
  • Point E has a positive y-coordinate.
    5·1 answer
  • HELP ASAP!!!!!!!!! Which sides form a right triangle?.
    7·1 answer
  • How are slope and derivatives related.
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!