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2 years ago

Help with the whole thing please and thanks

Mathematics

1 answer:

Mars2501 [29]2 years ago

6 0

1a. closest to 1

b. closest to 0

2a. closest to 1

b. closest to 1/2

Numb 3. Ain't clear.

4. Not sure.

5a. closer to 1

b.closer to 1/2

c. 3/4 - 5/12 = 1/3

6. 26/15

7. 7/18

8. 47/240

9. 191/225

10. 11/60

11. 81/200

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Another math two step equation, please help thank you.

ValentinkaMS [17]

1.25*(t)+7=43.25

To solve the 2nd one, you must do 43.25 - 7 = 26.25

36.25 divided by 1.25 = 29

4 0

3 years ago

28, 33, 88, 81, 40, 25, 41<br> Mean median mode and range

PolarNik [594]

Hello!

Step-by-step explanation:

Mean: 48

Median: 40

Mode: None

Range: 63

Hope this helps!

3 0

2 years ago

Read 2 more answers

Help help help if u can answer this I’ll mark u brainliest (6) (the one in the middle)

Hoochie [10]

You just have to plug in each thing:

So f(0):

3(0)^2 - 4 = -4

so its not A

f(-2) and f(2):

3(-2)^2 - 4

3(2)^2 -4

now we don't even have to calculate these, anything to the power of an even number is a positive number so -2^2 and 2^2 both equal 4

B is correct, so you can do the others if you want to check, but if B is true the others shouldn't be true.

4 0

3 years ago

Please help

maxonik [38]

Answer:

D. -12s squared + 11st - 2t squared

Step-by-step explanation:

FOIL:

-3s(4s) - 3s(-t) + 2t(4s) + 2t(-t)

- 12s² + 3st + 8st - 2t²

Simplify.

- 12s² + 11st - 2t²

7 0

3 years ago

Read 2 more answers

One model for the spread of a virusis that the rate of spread is proportional to the product of the fraction of the population P

Darya [45]

Answer:

The differential equation for the model is

$\frac{dP}{dt}=kP(1-P)$

The model for P is

$P(t)=\frac{1}{1-0.99e^{t/447}}$

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

Step-by-step explanation:

We can write the rate of spread of the virus as:

$\frac{dP}{dt}=kP(1-P)$

We know that P(0)=100 and P(3)=100+200=300.

We have to calculate t so that P(t)=0.9*100,000=90,000.

Solving the diferential equation

$\frac{dP}{dt}=kP(1-P)\\\\ \int \frac{dP}{P-P^2} =k\int dt\\\\-ln(1-\frac{1}{P})+C_1=kt\\\\1-\frac{1}{P}=Ce^{-kt}\\\\\frac{1}{P}=1-Ce^{-kt}\\\\P=\frac{1}{1-Ce^{-kt}}$

$P(0)= \frac{1}{1-Ce^{-kt}}=\frac{1}{1-C}=100\\\\1-C=0.01\\\\C=0.99\\\\\\P(3)= \frac{1}{1-0.99e^{-3k}}=300\\\\1-0.99e^{-3k}=\frac{1}{300}=0.99e^{-3k}=1-1/300=0.997\\\\e^{-3k}=0.997/0.99=1.007\\\\-3k=ln(1.007)=0.007\\\\k=-0.007/3=-0.00224=-1/447$

Then the model for the population infected at time t is:

$P(t)=\frac{1}{1-0.99e^{t/447}}$

Now, we can calculate t for P(t)=90,000

$P(t)=\frac{1}{1-0.99e^{t/447}}=90,000\\\\1-0.99e^{t/447}=1/90,000 \\\\0.99e^{t/447}=1-1/90,000=0.999988889\\\\e^{t/447}=1.010089787\\\\ t/447=ln(1.010089787)\\\\t=447ln(1.010089787)=447*0.010039225=4.487533$

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

8 0

3 years ago