What is 1+1??????????????????

Test the null hypothesis Upper H 0 : (mu 1 minus mu 2 )equals 0H0: μ1−μ2=0 versus the alternative hypothesis Upper H Subscript a

Law Incorporation [45]

Answer:

The test statistic t is t=2.9037.

The null hypothesis is rejected.

For a significance level of 0.05, there is enough evidence to support the alternative hypothesis.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>The sample 1, of size n1=25 has a mean of 1.15 and a standard deviation of 0.31. </em>

<em>The sample 2, of size n2=25 has a mean of 0.95 and a standard deviation of 0.15. </em>

This is a hypothesis test for the difference between populations means.

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

The significance level is α=0.05.

The difference between sample means is Md=0.2.

$M_d=M_1-M_2=1.15-0.95=0.2$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2+\sigma_2^2}{n}}=\sqrt{\dfrac{0.31^2+0.15^2}{25}}\\\\\\s_{M_d}=\sqrt{\dfrac{0.119}{25}}=\sqrt{0.005}=0.069$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.2-0}{0.069}=\dfrac{0.2}{0.069}=2.9037$

The degrees of freedom for this test are:

$df=n_1+n_2-1=25+25-2=48$

This test is a two-tailed test, with 48 degrees of freedom and t=2.9037, so the P-value for this test is calculated as (using a t-table):

$P-value=2\cdot P(t>2.9037)=0.0056$

As the P-value (0.0056) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

8 0

4 years ago

Explain why it can be helpful to use partial quotients when dividing?

Anastasy [175]

In mathematics, a quotient (from Latin: quotiens "how many times", pronounced ˈkwoʊʃənt) is the result of division. For example, when dividing 6 by 3, the quotient is 2, while 6 is called the dividend, and 3 the divisor.

3 0

3 years ago

Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2

Basile [38]

The Laplace transform of the given initial-value problem

$y' 5y = e^{4t}, y(0) = 2$ is mathematically given as

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is mathematically given as

$&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}$

$\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}$

$put $s=-s \Rightarrow a_{1}=\frac{17}{9}$$

$\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}$

In conclusion, Taking inverse Laplace tranoform

$L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\$

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

Read more about Laplace tranoform

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6 0

2 years ago

What are the values of x, y, and z in this system of equations?

scZoUnD [109]

Answer:

{x,y,z}={5,−3,3}

Step-by-step explanation:

[1] 2x + 4y + z = 1

[2] x - 2y - 3z = 2

[3] x + y - z = -1

// Solve equation [3] for the variable y

[3] y = -x + z - 1

// Plug this in for variable y in equation [1]

[1] 2x + 4•(-x +z -1) + z = 1

[1] -2x + 5z = 5

// Plug this in for variable y in equation [2]

[2] x - 2•(-x +z -1) - 3z = 2

[2] 3x - 5z = 0

// Solve equation [2] for the variable x

[2] 3x = 5z

[2] x = 5z/3

// Plug this in for variable x in equation [1]

[1] -2•(5z/3) + 5z = 5

[1] 5z/3 = 5

[1] 5z = 15

// Solve equation [1] for the variable z

[1] 5z = 15

[1] z = 3

// By now we know this much :

x = 5z/3

y = -x+z-1

z = 3

// Use the z value to solve for x

x = (5/3)(3) = 5

// Use the x and z values to solve for y

y = -(5)+(3)-1 = -3

Solution :

{x,y,z} = {5,-3,3}

5 0

3 years ago

What do the following two equations represent?

aleksandr82 [10.1K]

This is a graph of the two equations. This would then prove that the answer is d. None of the above. I’m unsure of this but Let me know if I’m correct

5 0

3 years ago