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olchik [2.2K]
3 years ago
5

Anfernee bought 4 umbrella and 5 hats and spent between $80.00 and 100.00. Each umbrella cost the same amount. Each hat cost the

same amount. The price of a hat is $4.00. What time s the least amount anfernee could have spent on an umbrella? What is the most anfernee most spent?
Mathematics
1 answer:
barxatty [35]3 years ago
6 0

Answer:

$15, $20

Step-by-step explanation:

Let u represent the unit price of umbrellas and h the unit price of hats.

Case 1:  A. spent $80.

Then $80 = 4u + 5($4), or $80 = $20 + 4u, or $60 = 4u, or

u = $15/umbrella.

Case 2:  A spent $100.

Then $100 = 4u + 5($4), or $100 = $20 + 4u, or $80 = 4u, or

u = $20/umbrella.

The least A. could have spent on a single umbrella was $15, and the most was $20.

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The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)
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Answer:

Cardiac output:F=0.055 L\s

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output:F=\frac{A}{\int\limits^T_0 {c(t)} \, dt} ---1

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\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt

Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}

Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

Hence Cardiac output:F=0.055 L\s

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