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Tcecarenko [31]

4 years ago

6

The Human Resources manager at your company reported that 2/5 of the employees have taken a sick day at least once this month.

1 answer:

almond37 [142]4 years ago

3 0

Wheres the rest of the question ?

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Is a hectare roughly equal to 2 tennis courts or to 2 soccer fields

lys-0071 [83]

It is most likely equal to two soccer fields

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3 years ago

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Why is convenience sampling biased? a. It takes too long to obtain b. none of the above c. The sample does not represent the pop

arlik [135]

Answer:

A

Step-by-step explanation:

5 0

3 years ago

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The graph of the function f(x) = ax2 + bx +C

morpeh [17]

Answer:

The axis of symmetry is -b/2a

The x-intercept shown is 3 units away from the axis of symmetry.

The other x-intercept also must be 3 units away from the axis of symmetry but on the opposite side.

Subtract 3 from the axis of symmetry.

The other x-intercept is (-b/2a-3,0)

Step-by-step explanation:

7 0

3 years ago

X^2+19=10 solve algebraically<br> 4x^2+3=43 solve algebraically

raketka [301]

The correct answer is B

8 0

3 years ago

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

3 0

4 years ago