Ask question

Ask question

All categories

3 years ago

8

What is the area of a triangle that has a base of 10 inches and a height of 7 inches

1 answer:

Natali [406]3 years ago

5 0

Answer:

it is 35

Step-by-step explanation:

You might be interested in

In the pattern above what will the 4th figure look like

Naddik [55]

Answer:

I don't know

Step-by-step explanation:

because I don't see the pattern

4 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

Question is on the bottom please help me

adell [148]

I got ~~28.3 hope this helps :)

7 0

3 years ago

We define x as a year between 2008 and 2013 and y as the total number of smartphones sold that year, in

stepan [7]

17.3 million smartphones were sold in 2009

4 0

3 years ago

Read 2 more answers

You are given the following information obtained from a random sample of 5 observations. 20 18 17 22 18 At 90% confidence, you w

Margaret [11]

Answer:

a

The null hypothesis is

$H_o : \mu = 21$

The Alternative hypothesis is

$H_a : \mu< 21$

b

$\sigma_{\= x} = 0.8944$

c

$t = -2.236$

d

Yes the mean population is significantly less than 21.

Step-by-step explanation:

From the question we are given

a set of data

20 18 17 22 18

The confidence level is 90%

The sample size is n = 5

Generally the mean of the sample is mathematically evaluated as

$\= x = \frac{20 + 18 + 17 + 22 + 18}{5}$

$\= x = 19$

The standard deviation is evaluated as

$\sigma = \sqrt{ \frac{\sum (x_i - \= x)^2}{n} }$

$\sigma = \sqrt{ \frac{ ( 20- 19 )^2 + ( 18- 19 )^2 +( 17- 19 )^2 +( 22- 19 )^2 +( 18- 19 )^2 }{5} }$

$\sigma = 2$

Now the confidence level is given as 90 % hence the level of significance can be evaluated as

$\alpha = 100 - 90$

$\alpha = 10$ %

$\alpha =0.10$

Now the null hypothesis is

$H_o : \mu = 21$

the Alternative hypothesis is

$H_a : \mu< 21$

The standard error of mean is mathematically evaluated as

$\sigma_{\= x} = \frac{\sigma}{ \sqrt{n} }$

substituting values

$\sigma_{\= x} = \frac{2}{ \sqrt{5 } }$

$\sigma_{\= x} = 0.8944$

The test statistic is evaluated as

$t = \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }$

substituting values

$t = \frac{ 19 - 21 }{ 0.8944 }$

$t = -2.236$

The critical value of the level of significance is obtained from the critical value table for z values as

$z_{0.10} = 1.28$

Looking at the obtained value we see that $z_{0.10}$ is greater than the test statistics value so the null hypothesis is rejected

6 0

4 years ago