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3 years ago

How is solving an inequality similar to solving an equation?

Mathematics

1 answer:

LenaWriter [7]3 years ago

5 0

I think they are similar because they are both solving for a certain number (sum, product quotient)

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Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

2 years ago

Hello can someone help me with this

Degger [83]

Answer:

I am going to answer the ones you haven't answered yet...

Bullet point 2: The Y-intercept is where the graph crosses the y-axis in this case that would be y=0 or (0,0)

Bullet Point 5: the question basically says that for the values of x when they are between 0 and 6 the y values are ___. the answer for this one would be y ≥ 0.

5 0

1 year ago

Read 2 more answers

The sum of two numbers is 48.when the smaller number is subtracted from the larger number,the difference is 12. What are the two

Savatey [412]

Answer:

The two numbers are:

30 and 18

Step-by-step explanation:

a + b = 48

a - b = 12

From the first eq.

a = 48 - b

Replacing this last value in the second eq.

(48 - b) - b = 12

48 - 2b = 12

48 - 12 = 2b

36 = 2b

36/2 = b

18 = b

from the first eq.

a + 18 = 48

a = 48 - 18

a = 30

Check:

from the second eq.

a - b = 12

30 - 18 = 12

5 0

2 years ago

The Problem of the Points. Two players engage in a game. They are equally matched, and the chances of winning a point are the sa

Dominik [7]

Answer:

Player A receives $77.34375 of the sakes

Player B receives $22.65625 of the stakes

Step-by-step explanation:

Given that at the time the game was interrupted, we have;

The number of points player A has = 7 points

The number of points player B has = 5 points

The amount the first player to score 10 points wins = $100

The number of points remaining for player A to win the game = 3

The number of points remaining for player B to win the game = 5

Therefore, the number of trials remaining for a winner to emerge = 3 + 5 - 1 = 7 trials

We take the probability that Player A wins a point as success

We find the likelihood of Plater A winning by using binomial theory as follows;

$P(S) = \sum\limits_{j=3}^7 \dbinom{7}{3} \cdot \dfrac{1}{2^7} = \dfrac{35}{128} + \dfrac{35}{128} + \dfrac{21}{128} + \dfrac{7}{128} + \dfrac{1}{128} = \dfrac{99}{128}$

Therefore, given that the likelihood of Player A winning = 99/128

The stakes should be divided such that Player A gets 99/128 share of the stakes while Player B gets 1 - 99/128 = 29/128 share of the stakes

Therefore, Player A gets (99/128) × $100 = $77.34375

Player B gets (29/128) × $100 = $22.65625

8 0

2 years ago

A bag of sugar contains 420g. A

faltersainse [42]

Answer:

63g extra

Step-by-step explanation:

420 x 0.15 =63 g

6 0

2 years ago