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KIM [24]
3 years ago
14

You pick a card and a marble randomly.

Mathematics
1 answer:
Iteru [2.4K]3 years ago
7 0

Answer:

B2/25

Step-by-step explanation:

because that it is the propability

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The recycling center processed 36,000 pounds of recyclable materials. How many tons of recyclable materials are processed?
olchik [2.2K]

Answer:

18 Tons

Step-by-step explanation:

36,000 / 2000 = 18

3 0
3 years ago
Share £40 in the ratio 1:4 between Tim and Sam
lesantik [10]

Answer:

\£8:\£32

Step-by-step explanation:

\frac{40}{1+4}

\frac{40}{5}

=8

1:4

1 \times 8:4 \times 8

8:32

7 0
3 years ago
Read 2 more answers
The length of a rectangle is 4 inches more than 4 times the width. The perimeter is 138 inches. Find the length and the width.
MatroZZZ [7]
4 times 4 which is 8 so 8 times 4 which is 32 so divide that by 138 but divide the 138 first and you will get 17.25?
5 0
3 years ago
Hi, I have utterly no idea what I'm doing here.
Anna11 [10]

Answer:

Factored form: 2x(2x^2+x+1)

Step-by-step explanation:

You're subtracting them. :)

6 0
3 years ago
Read 2 more answers
A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so
Debora [2.8K]

Let A(t) = amount of salt (in pounds) in the tank at time t (in minutes). Then A(0) = 11.

Salt flows in at a rate

\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}

and flows out at a rate

\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}

which I'll solve with the integrating factor method.

\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95

-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}

\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}

Integrate both sides. By the fundamental theorem of calculus,

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)

\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}

\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}

\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}

After 1 hour = 60 minutes, the tank will contain

A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131

pounds of salt.

7 0
2 years ago
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