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3 years ago

1 a. 10k + 8(3 + 9k) + 9

Mathematics

2 answers:

viva [34]3 years ago

8 0

Answer:

The answer is 82k +33

Step-by-step explanation:

10k + 8(9k+3) +9

adell [148]3 years ago

6 0

Hello User!

Answer: 82k+33

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Tickets for a softball game at EAHS cost $5 for adults and $1 for students. The attendance that

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3 years ago

The differential equation in Example 3 of Section 2.1 is a well-known population model. Suppose the DE is changed to dP dt = P(a

LuckyWell [14K]

Answer:

Decreases

Step-by-step explanation:

We need to determine the integral of the DE;

$dP/dt=P(aP-b)$

$dP=P(aP-b)dt$

$1/(dP^2-bP)dP=dt$

We can solve this by integration by parts on the left side. We expand the fraction 1/P²:

$1/(d-b/P)\cdot{P^2} dP$

let

$u=d-b/P$

$du/dP=b/P^2$

$dP=$ $\int\limits {P^2/b} \, du$

$P=lnu/b$

Substitute u in:

$P=ln(d-b/P)/b$

Therefore the equation is:

$ln(d-b/P)/b=t$

We simplify:

$d-b/P=e^b^t$

$P=b/(d-e^b^t)$

As t increases to infinity P will decrease

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3 years ago

Convert 1111+102 into quinary number

Kipish [7]

Answer:

13421+402

Step-by-step explanation:

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3 years ago

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typo

muminat

Answer:

The required probability is 0.55404.

Step-by-step explanation:

Consider the provided information.

The number of typographical errors on a page of the first booklet is a Poisson random variable with mean 0.2. The number of typographical errors on a page of second booklet is a Poisson random variable with mean 0.3.

Average error for 7 pages booklet and 5 pages booklet series is:

λ = 0.2×7 + 0.3×5 = 2.9

According to Poisson distribution: $P(k{\text{ events in interval}})={\frac {\lambda ^{k}e^{-\lambda }}{k!}}$

Where $\lambda$ is average number of events.

The probability of more than 2 typographical errors in the two booklets in total is:

$P(k > 2)= 1 - {P(k = 0) + P(k = 1) + P(k = 2)}$

Substitute the respective values in the above formula.

$P(k > 2)= 1 - ({\frac {2.9 ^{0}e^{-2.9}}{0!}} + \frac {2.9 ^{1}e^{-2.9}}{1!}} + \frac {2.9 ^{2}e^{-2.9}}{2!}})$

$P(k > 2)= 1 - (0.44596)$

$P(k > 2)=0.55404$

Hence, the required probability is 0.55404.

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3 years ago

A recipe for cookies requires 1 3/4 cups of sugar

bekas [8.4K]

9.625 cups of sugar. just times 1.75(1 3/4) by 5.5 so that's gonna give you an answer of 9.625 cups of sugar

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