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nydimaria [60]
1 year ago
5

A project’s critical path is composed of activities A (variance .33), B (variance .67), C (variance .33), and D (variance .17).

What is the standard deviation on the critical path?
Mathematics
1 answer:
julia-pushkina [17]1 year ago
4 0

Answer:

The standard deviation on the critical path = 0.834

Step-by-step explanation:

Variance of activity A (V_{1}) = 0.33

Variance of activity B (V_{2}) = 0.67

Variance of activity C (V_{3}) = 0.33

Variance of activity D (V_{4}) = 0.17

Thus the standard deviation on the critical path (\alpha ^{2}) = V_{1} ^{2} + V^2_{2} +  V^2_{3} + V^2_{4}

                                                                             \alpha ^{2}  = 0.33^{2} + 0.67^{2} + 0.33^{2} + 0.17^{2}

                                                                             \alpha ^{2} = 0.6956

                                                                             \alpha = 0.834

The standard deviation on the critical path = 0.834

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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)
Lemur [1.5K]

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

A = 3 mg

\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

4 0
1 year ago
Your cell phone plan charges a base charge each month plus a charge per daytime minute of usage and December used 510 daytime mi
algol13

Answer:

In February, 423 daytime minutes is used

Step-by-step explanation:

Let the base plan charges be x

And cost per daytime minute be y

In December,

x  + 510y =  92.25------------------(1)

In January,

x + 397y = 77.56---------------------(2)

Subtracting eq(2) from eq(1)

x  + 510y =  92.25

x + 397y = 77.56

-------------------------------

0  + 113y = 14.69

-------------------------------

y = \frac{14.69}{113}

y =  0.13----------------------------------(3)

Substituting (3) in (1)

x  + 510(0.13) =  92.25

x + 66.3 = 92.25

x = 92.25 - 66.3

x = 25.95

So In February

base plan  + (daytime minute)(cost per daytime minute) = 80.9

25.95 + (daytime minute)(0.13) = 80.9

(daytime minute)(0.13) = 80.9 - 25.95

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(daytime minute) =\frac{54.95}{0.13}

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daytime minute  \approx 423

3 0
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