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Viefleur [7K]
1 year ago
9

The length of time Y necessary to complete a key operation in the construction of houses has an exponential distribution with me

an 13 hours. The formula C = 100 + 80Y + 3Y2 relates the cost C of completing this operation to the square of the time to completion. The mean of C was found to be found to be 2,154 hours and the variance of C was found to be 10,440,820.
Mathematics
1 answer:
Angelina_Jolie [31]1 year ago
5 0

Answer:

Reference The length of time Y necessary to. ... construction of houses has an exponential distribution with mean 10 hours. The formula C = 100 + 40Y + 3Y 2 relates the cost C of completing this operation to ... Find the mean and variance of C. ... The length of time necessary to complete a key operation in the construction of ...

Step-by-step explanation:

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In some languages, every consonant must be followed by a vowel. How many seven-letter ‟words” can be made from the Hawaiian word
emmasim [6.3K]

Answer:

One. Kamala.

Step-by-step explanation:

7 0
1 year ago
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
Lin counts 5 bacteria under a microscope. She counts them again each day for four days, and finds
svetlana [45]

Answer:

Yes, but it is not linear. It is exponential :  f(x) = 5\cdot 2^{x-1}

Step-by-step explanation:

On the first day we have 5 bacteria.

On the second day we will have 5*2 = 10 bacteria.

On the third day we will have 10*2 = 5*2*2 = 5*2^2 = 20 bacteria.

On the fourth day we will have 20*2 = 5*2*2*2 = 5*2^3 = 40 bacteria.

We can see that

                                                f(x) = 5\cdot 2^{x-1},

where x is a number of days and f(x) gives us the number of bacteria.

4 0
1 year ago
For which discriminant is the graph possible<br> b2-4ac=0<br><br> b2-4ac=-1<br><br> b2-4ac=4
VARVARA [1.3K]

Answer:

The graph is possible for b^2-4ac=4

Step-by-step explanation:

we know that

The discriminant of a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

D=b^2-4ac

If D=0 the quadratic equation has only one real solution

If D>0 the quadratic equation has two real solutions

If D<0 the quadratic equation has no real solution (complex solutions)

In this problem , looking at the graph, the quadratic equation has two real solutions (the solutions are the x-intercepts)

so

b^2-4ac > 0

therefore

The graph is possible for b^2-4ac=4

4 0
1 year ago
AX and EX are secant segments that intersect at point X. What is the length of DE?
vazorg [7]

Answer:

we kind of need to see the diagram

Step-by-step explanation:

5 0
1 year ago
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