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PSYCHO15rus [73]

3 years ago

12

a story has a contest to guess the mass of 10000 peanuts. If a peanut has a mass of about 0.45 then what would be a reasonable g

uess for the mass of 10000

2 answers:

mamaluj [8]3 years ago

7 0

1000000 i think...................................................

Alex_Xolod [135]3 years ago

7 0

Around 4,500 because .45 times 10000 = 4,500

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3 years ago

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

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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

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

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

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

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

$P(0) = 100$

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

$t = 10$

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

$t = 20$

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

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

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:

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

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

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

$P(\infty) = 1200$

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What is the HCF of 60 and 96?

pochemuha

Answer:
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Explanation:
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The population in 1993 was 11,211 people. The annual increase of the population is 4.18%. What is the population of Cypress in 2

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Answer:

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Step-by-step explanation:

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4 years ago

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Eight more than the product of two and a number x

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