Find the perimeter and area of the figure below.

A smart phone manufacturer wants to determine the average battery time that it’s new smart phone will get. The manufacturer will

sasho [114]

Answer:

Note: <em>The full question is attached as picture below</em>

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1. The manufacturer is taking a sample of her friends, which is most convenient to sample - Convenience sampling

2. The whole population is divided into different groups (strata) and random samples are drawn from each group - Stratified sampling

3. The population is divided into different clusters (here, communities) and whole clusters are surveyed - Cluster sampling

4. The samples are drawn at a fixed interval from the population list - Systematic sampling

5. The samples are randomly drawn from the population - Simple random sampling

7 0

3 years ago

What’s the median of 82 86 90 94 88 76 74 74 92 96

Galina-37 [17]

Answer:

88 and 76 are the median!

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

On the Sat the lowest value anyone can score on the math section is 200 and the highest is 800. Jason is a senior retaking the S

schepotkina [342]

Answer:

10 people

Step-by-step explanation:

Parking needs to be paid for but if all friends are going together, they need to only pay once for parking.

The amount spent on just tickets is therefore:

= 243.25 - 15.75

= $227.50.

If $227.50 was for tickets alone, the number of tickets they could buy is:

= Total amount left/ Price of tickets

= 227.50 / 22.75

= 10 people

Step-by-step explanation:

6 0

2 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

3 years ago

Last year at a certain high school, there were 50 boys on the honor roll and 75 girls on the honor roll. This year, the number o

Arturiano [62]

Answer:

About 15.3%!

Step-by-step explanation:

ill explain in comments, i hope this helps you out

6 0

3 years ago

Read 2 more answers