All categories

3 years ago

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

You might be interested in

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