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ivanzaharov [21]

3 years ago

10

A carpenter can install 8 sheets of drywall in 10 minutes. At this rate, how long does it take to install one sheet of drywall?

How long will it take the carpenter to install 60 sheets?

1 answer:

dlinn [17]3 years ago

5 0

Answer:

75 seconds for one sheet of drywall

75 minutes to install 60 sheets

Step-by-step explanation:

10 minutes / 8 sheets = 1.25 minutes per sheet (minutes/sheet) = 75 seconds

60 sheets * 1.25 minutes per sheet = 75 minutes

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A tire company finds that the lifespan for one brand of its tires is normally distributed with a mean of 47,500 miles and a stan

Illusion [34]

Answer:

42580 miles

Step-by-step explanation:

Mean = $\mu = 47500$

$\sigma = 3000$

The manufacturer does not want to replace more than 5% of the tires

$P(X\leq x)=5\%$

$P(\frac{x-\mu}{\sigma}\leq \frac{x-47500}{3000})=0.05$

By using normal table values :

$\frac{x-47500}{3000}=-1.64$

$x=(-1.64 \times 3000)+47500$

$x=42580$

Hence the approximate number of miles for the warranty is 42580 miles

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3 years ago

A local fast-food restaurant can serve

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Answer:

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3 years ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

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3 years ago

What is h(-4) for h(x) =(x-2)^2 +3x

PSYCHO15rus [73]

$(- 4 - 2)^{2} + 3( - 4) \\ = ( - 6)^{2} + ( - 12) \\ = 36 - 12 \\ = 24$
The answer is 24

7 0

2 years ago

which number is one tenth of the expanded fom below? (3×1000)+(2×100)+(7×10)+(5×1)+(2×0.1)+(5×0.010) A 326.525. B 3,265.25. C 32

solong [7]

B
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4 0

3 years ago

Read 2 more answers