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kirza4 [7]
3 years ago
5

Mrs. Schmidt gave her students the list of numbers shown below.

Mathematics
2 answers:
Rina8888 [55]3 years ago
6 0

Answer:

Convert the words to decimals then it will be easy

Step-by-step explanation:

lorasvet [3.4K]3 years ago
6 0

Answer:

Option A: -4.50,  -\frac{15}{16}, -1\frac{8}{11}

Step-by-step explanation:

Draw a number line and mark the point -0.58, anything on the left of the point is less than -0.58.

=> <u>-4.50 < -0.58</u>

=>  -\frac{15}{16} = -0.94,<u> -0.94 < -0.58</u>

=> -1\frac{8}{11} = -1.72, <u>-1.72 < -0.58</u>

Hope this helps!

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Suppose a certain population satisfies the logistic equation given by dP
Ksenya-84 [330]

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The population when t = 3 is 10.

Step-by-step explanation:

Suppose a certain population satisfies the logistic equation given by

\frac{dP}{dt}=10P-P^2

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Using variable separable method we get

\frac{dP}{10P-P^2}=dt

Integrate both sides.

\int \frac{dP}{10P-P^2}=\int dt             .... (1)

Using partial fraction

\frac{1}{P(10-P)}=\frac{A}{P}+\frac{B}{(10-P)}

A=\frac{1}{10},B=\frac{1}{10}

Using these values the equation (1) can be written as

\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt

\int \frac{dP}{10P}+\int \frac{dP}{10(10-P)}=\int dt

On simplification we get

\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C

\frac{1}{10}(\ln \frac{P}{10-P})=t+C

We have P(0)=1

Substitute t=0 and P=1 in above equation.

\frac{1}{10}(\ln \frac{1}{10-1})=0+C

\frac{1}{10}(\ln \frac{1}{9})=C

The required equation is

\frac{1}{10}(\ln \frac{P}{10-P})=t+\frac{1}{10}(\ln \frac{1}{9})

Multiply both sides by 10.

\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}

e^{\ln \frac{P}{10-P}}=e^{10t+\ln \frac{1}{9}}

\frac{P}{10-P}=\frac{1}{9}e^{10t}

Reciprocal it

\dfrac{10-P}{P}=9e^{-10t}

P(t)=\dfrac{10}{1+9e^{-10t}}

The population when t = 3 is

P(3)=\dfrac{10}{1+9e^{-10\cdot 3}}

Using calculator,

P=9.999\approx 10

Therefore, the population when t = 3 is 10.

8 0
3 years ago
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