All categories

3 years ago

Will someone slove this please

Mathematics

1 answer:

Evgesh-ka [11]3 years ago

4 0

C. I think is the answer

You might be interested in

Of 960 pages 384 had an add. What's the probability it contains an add

kramer

384/960
Thus, answer=2/5

4 0

3 years ago

What is the equation of the line in point-slope form that has a slope of -3 and passes through the point (5, 9)?

krek1111 [17]

Answer:

$y-9=-3(x-5)$

Step-by-step explanation:

The equation of the line in point-slope form is given as

$y-y_1=m(x-x_1)$

Here m=-3 and (x1, y1)=(5, 9)

Therefore,

$y-9=-3(x-5)$

8 0

3 years ago

Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that

amm1812

Answer:

The value is $P(| \^ p - p| < 0.05 ) = 0.9822$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.52$

The sample size is n = 563

Generally the population mean of the sampling distribution is mathematically represented as

$\mu_{x} = p = 0.52$

Generally the standard deviation of the sampling distribution is mathematically evaluated as

$\sigma = \sqrt{\frac{ p(1- p)}{n} }$

=> $\sigma = \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }$

=> $\sigma = 0.02106$

Generally the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

$P(| \^ p - p| < 0.05 ) = P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 ))$

Here $\^ p$ is the sample proportion of persons with a college degree.

$P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma } < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )$

Here

$\frac{[\^p - p] - p}{\sigma } = Z (The\ standardized \ value \ of\ (\^ p - p))$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 } < Z < \frac{-0.47 + 0.52}{0.02106 }]$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P[ -2.37 < Z < 2.37 ]$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(Z < 2.37 ) - P(Z < -2.37 )$

From the z-table the probability of (Z < 2.37 ) and (Z < -2.37 ) is

$P(Z < 2.37 ) = 0.9911$

and

$P(Z < - 2.37 ) = 0.0089$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) =0.9911-0.0089$

=> $P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = 0.9822$

=> $P(| \^ p - p| < 0.05 ) = 0.9822$

3 0

3 years ago

Solve the following differential equation using using characteristic equation using Laplace Transform i. ii y" +y sin 2t, y(0) 2

kifflom [539]

Answer:

The solution of the differential equation is $y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

<u>i) Using characteristic equation:</u>

The characteristic equation method assumes that y(t)= $e^{rt}$ , where "r" is a constant.

We find the solution of the homogeneus differential equation:

y" + y = 0

$y'=re^{rt}$

$y"=r^{2}e^{rt}$

$r^{2}e^{rt}+e^{rt}=0$

$(r^{2}+1)e^{rt}=0$

As $e^{rt}$ could never be zero, the term (r²+1) must be zero:

(r²+1)=0

r=±i

The solution of the homogeneus differential equation is:

$y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}$

Using Euler's formula:

$y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]$

$y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)$

$y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)$

The particular solution of the differential equation is given by:

$y(t)_{p}=ASin(2t)+BCos(2t)$

$y'(t)_{p}=2ACos(2t)-2BSin(2t)$

$y''(t)_{p}=-4ASin(2t)-4BCos(2t)$

So we use these derivatives in the differential equation:

$-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)$

$-3ASin(2t)-3BCos(2t)=Sin(2t)$

As there is not a term for Cos(2t), B is equal to 0.

So the value A=-1/3

The solution is the sum of the particular function and the homogeneous function:

$y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)$

Using the initial conditions we can check that C1=5/3 and C2=2

<u>ii) Using Laplace Transform:</u>

To solve the differential equation we use the Laplace transformation in both members:

ℒ[y" + y]=ℒ[Sin(2t)]

ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1

ℒ[y]=Y(s)

ℒ[Sin(2t)]= $\frac{2}{(s^{2}+4)}$

We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) = $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)-2s-1= $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)= $\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}$

Y(s)= $\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}$

Y(s)= $\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}$

Using partial franction method:

$\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}$

2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)

We solve the equation system:

A+C=2

B+D=1

A+4C=8

B+4D=6

The solutions are:

A=0 ; B= -2/3 ; C=2 ; D=5/3

So,

Y(s)= $\frac{-\frac{2}{3} }{s^{2}+4} +\frac{2s+\frac{5}{3} }{s^{2}+1}$

Y(s)= $-\frac{1}{3} \frac{2}{s^{2}+4} +2\frac{s }{s^{2}+1}+\frac{5}{3}\frac{1}{s^{2}+1}$

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[ $-\frac{1}{3} \frac{2}{s^{2}+4}$ ]-ℒ⁻¹[ $2\frac{s }{s^{2}+1}$ ]+ℒ⁻¹[ $\frac{5}{3}\frac{1}{s^{2}+1}$ ]

$y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

3 0

3 years ago

2. Carly needed to study for of an hour for 9 days. How many hours did she study?

Leya [2.2K]

Answer: The closest answer I got is 9 5/6 hours

5 0

2 years ago