5 pounds equals how many ounces

Write Two Different Addition Equations That Have 12 As The Solution.

kow [346]

6+6=12
8+4=12

or if you mean algebraic equation:

6+x=12
x+4=12

5 0

4 years ago

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

The J.R. Ryland Computer Company is considering a plant expansion to enable the company to begin production of a new computer pr

Solnce55 [7]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data:

Medium-Scale Large-Scale

Expansion Profit Expansion Profit

x f(x) y f(y)

Low 50 0.2 0 0.2

Demand Medium 150 0.5 100 0.5

High 200 0.3 300 0.3

a. Compute the expected value for the profit associated with the two expansion alternatives.

Which decision is preferred for the objective of maximizing the expected profit?

Expected value for medium scale expansion profit :

Expected value (E) = Σ(X) * f(x)

Σ[(50 * 0.2) + (150 * 0.5) + (200 * 0.3)]

= 145

Expected value for Large scale expansion profit :

Expected value (E) = Σ(X) * f(x)

Σ[(0 * 0.2) + (100 * 0.5) + (300 * 0.3)]

= 140

Medium scale expansion profit is preferred as it has the highest expected value.

b. Compute the variance for the profit associated with the two expansion alternatives.

Which decision is preferred for the objective of minimizing the risk or uncertainty?

Variance (V) = Σ(X - E)² * f(x):

Variance for medium scale expansion profit :

V = [((50-145)^2 * 0.2) + ((150-145)^2 * 0.5) + ((200-145)^2 * 0.3) = 2725

Variance for Large scale expansion profit :

V = [((0-140)^2 * 0.2) + ((100-140)^2 * 0.5) + ((300-140)^2 * 0.3) = 12400

Smaller variance is required to minimize risk, Hence, choose the medium scale expansion profit.

3 0

3 years ago

How many people will 5 pitchers serve if 1/8 pitchers served one person?

Juliette [100K]

5 pitchers * 8 people per pitcher= 40 people

7 0

4 years ago

Read 2 more answers

What is 1/10 of 0.04

OleMash [197]

The answer is .004

Hope It Helps!

5 0

3 years ago

Read 2 more answers