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aleksklad [387]
3 years ago
10

Convert 198%, as fraction and decimal​

Mathematics
2 answers:
jarptica [38.1K]3 years ago
6 0

Answer:1.98 and 1 49/50

Step-by-step explanation:

Alenkasestr [34]3 years ago
6 0

Answer:

Step-by-step explanation:

198% = 198/100 = 99/50 = 1.98

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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
What is the equation when y=5x+4 is reflected over the x-axis, y-axis & y=x?
tigry1 [53]

Answer:

A) y=-5x-4

B) y=-5x+4

C) y=\frac{x-4}{5}

Step-by-step explanation:

So we have the equation:

y=5x+4

Let's write this in function notation. Thus:

y=f(x)=5x+4

A)

To flip a function over the x-axis, multiply the function by -1. Thus:

f(x)=5x+4\\-(f(x))=-(5x+4)

Simplify:

-f(x)=-5x-4

B) To flip a function over the y-axis, change the variable x to -x. Thus:

f(x)=5x+4\\f(-x)=5(-x)+4

Simplify:

f(-x)=-5x+4

C) A reflection over the line y=x is synonymous with finding the inverse of the function.

To find the inverse, switch x and f(x) and solve for f(x):

f(x)=5x+4

Switch:

x=5f^{-1}(x)+4

Subtract 4 from both sides:

x-4=5f^{-1}(x)

Divide both sides by 5:

f^{-1}(x)=\frac{x-4}{5}

And we're done :)

7 0
3 years ago
Express {x+1}{2x-3} as a trinomial.
Marizza181 [45]

Answer:

2x^2 + 5x + 3

Step-by-step explanation:

First, expand the original equation:

(x + 1)(2x + 3)

2x^2 + 3x + 2x + 3

Now simplify:

2x^2 + 5x + 3

And that's your answer!

6 0
3 years ago
What is value of x?<br><br><br><br> Enter your answer in the box.<br><br> x =
Ganezh [65]

Answer:

I belive x=4.9

Step-by-step explanation:

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3 years ago
Evaluate the product: (a-10)*(a-9).....(a-1)*a
lilavasa [31]

Answer:

10x9-5(8)=34

Step-by-step explanation:

6 0
3 years ago
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