Change the following quantity into the indicated unit.

In each of problems 5 through 11, find the general solution of the given differential equation

Brrunno [24]

The complete question is

"Find the general solution of the given differential equation

y''-y=0, y1(t)=e^t , y2(t)=cosht

The function $y(t)=e^t$ is the solution of the given differential equation.

The function y(t)=cosht is the solution of given differential equation.

<h3>What is a function?</h3>

The function is a type of relation, or rule, that maps one input to specific single output.

Given;

$y_1(t) = e^t$

Given differential equations are,

y''-y = 0

So that,

$y' (t) = e^t, y'' (t) = e^t$

Substitute values in the given differential equation.

$e^t -e^t=0$

Therefore, the function $y(t)=e^t$ is the solution of the given differential equation.

Another function;

$y(t)=cosht$

So that,

$y"(t)=sinht\\\\y"(t)=cosht$

Hence, function y(t)=cosht is solution of given differential equation.

Learn more about function here:

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7 0

2 years ago

The realtor and her clients do not know the average home sale price for all of Guelph (500 was actually just a guess). However,

strojnjashka [21]

Answer:

a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=400$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=80$ represent the population standard deviation

n=15 represent the sample size

2) Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

Now we have everything in order to replace into formula (1):

$400-2.58\frac{80}{\sqrt{15}}=346.708$

$400+2.58\frac{80}{\sqrt{15}}=453.292$

So on this case the 99% confidence interval would be given by (346.708;453.292)

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

We need to find the degrees of freedom first df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that $t_{\alpha/2}=2.98$