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3 years ago

Calculate the coefficient of x4y18 in the expansion of (x − y)22

1 answer:

4 0

Binomial theorem:

$(x-y)^n=\displaystyle\sum_{k=0}^n\binom nkx^k(-y)^{n-k}$

The term with $x^4y^{18}$ corresponds to $n=18+4=22$ and $k=4$ (or $k=18$ , since $\dbinom nk=\dbinom n{n-k}$ ).

So, the coefficient of this term is

$\dbinom{22}4(-1)^{18}=\dbinom{22}4=\dfrac{22!}{4!(22-4)!}=7315$

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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

3 years ago

Halp me pls..<br> !!!!<br><br><br> :D I will be happy

pochemuha

The probability of multiple events happening is found by multiplying the probabilities of each event together.

$P_{d+even\ number}=P_{d}*P_{even}\\\\P_{d}=\frac{number\ of\ events}{total\ number\ of\ events}=\frac{(d)}{a,\ b,\ c,\ d,\ e)}=\frac{1}{5}\\\\P_{even}=\frac{number\ of\ events}{total\ number\ of\ events}=\frac{(2,\ 4,\ 6)}{(1,\ 2,\ 3,\ 4,\ 5,\ 6)}=\frac{3}{6}=\frac{1}{2}\\\\\frac{1}{5}*\frac{1}{2}=\boxed{\frac{1}{10}}$

So yes, 1/10 is the answer :)

5 0

3 years ago

Which equation requires the division property of equality to be solved?

Ymorist [56]

Answer:

A. 201.6 = 11.2f

Step-by-step explanation:

we solve for each unknown to verify which equality will require the division property:

#A. 201.6 = 11.2f

$201.6=11.2f\\\\f=\frac{201.6}{11.2}=201.6\div 11.2\\\\=18$