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xenn [34]
3 years ago
5

Sylvia is considering investing in one of two bond packages offered to her by different brokers. Broker U suggests that Sylvia b

uy two par value $1,000 bonds from Franklin County, three par value $500 bonds from Enam Telecom, and two par value $1,000 bonds from the city of Iligs.. Franklin County bonds are selling at 96.674, Enam Telecom bonds are selling at 109.330, and Iligs bonds are selling at 103.851. Broker V suggests that Sylvia buy four par value $500 bonds from Trochel Office Supplies, one par value $500 bond from Okaloosa county, and three par value $1,000 bonds from Globin Publishing. Bonds from Trochel Office Supplies are selling at 105.142, Okaloosa county bonds are selling at 85.990, and Globin Publishing bonds are selling at 97.063. If Broker U charges a commission of 2.8% of the market value of the bonds sold and Broker V charges a fee of $65 for each bond sold, which bond package will cost Sylvia less, and by how much? a. Broker V’s bond package will cost Sylvia $366.00 less than Broker U’s. b. Broker V’s bond package will cost Sylvia $205.77 less than Broker U’s. c. Broker U’s bond package will cost Sylvia $156.02 less than Broker V’s. d. Broker U’s bond package will cost Sylvia $361.79 less than Broker V’s.
Mathematics
2 answers:
Aneli [31]3 years ago
8 0
Thanks for your question, Broker U's package will cost her $156.02 less than broker V's.
Ket [755]3 years ago
6 0

Answer:

Based on what was said by the question, Broker U's package will cost her $156.02 less than broker V's.

Step-by-step explanation:

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A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan
Artyom0805 [142]

Answer:

1) \frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let y represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: \frac{dy}{dt}=rate\:in-rate\:out

The amount coming in is 0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}

The rate going out depends on the concentration of salt in the tank at time t.

If there is y(t) pounds of  salt and there are 100+2t gallons at time t, then the concentration is: \frac{y(t)}{2t+100}

The rate of liquid leaving is is 3gal\min, so rate out is =\frac{3y(t)}{2t+100}

The required differential equation becomes:

\frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) We rewrite to obtain:

\frac{dy}{dt}+\frac{3}{2t+100}y=2.5

We multiply through by the integrating factor: e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }

to get:

(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }

This gives us:

((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }

We integrate both sides with respect to t to get:

(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C

Multiply through by: (50+t)^{-\frac{3}{2}} to get:

y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }

y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}

We apply the initial condition: y(0)=0

0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}

C=-12500\sqrt{2}

The amount of salt in the tank at time t is:

y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}

y(50)=98.23

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})

As t\to \infty, 50+t\to \infty and \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0

This implies that:

\lim_{t \to \infty}y(t)=\infty- 0=\infty

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

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Answer:

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