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vampirchik [111]
3 years ago
10

Name all the awnsers below

Mathematics
1 answer:
spayn [35]3 years ago
4 0

Answer:

this doesn't look fun.

Step-by-step explanation:

You might be interested in
Watching tv: in 2012, the general social survey asked a sample of 1310 people how much time they spend\t watching tv each day. T
garik1379 [7]

Answer:

a) Null hypothesis:\mu \geq 3  

Alternative hypothesis:\mu < 3  

This hypothesis test is a left tailed test.

b) t=\frac{2.8-3}{\frac{2.6}{\sqrt{1310}}}=-2.784  

The p value for this case can be calculated with this probability:  

p_v =P(z  

We can conduct the test with the Ti84 using the following steps:

STAT>TESTS>T-test>Stats

We input the value \mu_o =3, \bar X= 2.8, s_x = 2.6, n=1310 and for the alternative we select < \mu_o. Then press Calculate.

And we got the same results.  

Step-by-step explanation:

Information given

\bar X=2.8 represent the sample mean

s=2.6 represent the population standard deviation  

n=1310 sample size  

\mu_o =3 represent the value to test

\alpha=0.5 represent the significance level for the hypothesis test.  

t would represent the statistic

p_v represent the p value for the test

Part a) System of hypothesis

We want to test if the true mean is less than 3, the system of hypothesis would be:  

Null hypothesis:\mu \geq 3  

Alternative hypothesis:\mu < 3  

This hypothesis test is a left tailed test.

Part b

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Replacing the info given we got:

t=\frac{2.8-3}{\frac{2.6}{\sqrt{1310}}}=-2.784  

The p value for this case can be calculated with this probability:  

p_v =P(z  

We can conduct the test with the Ti84 using the following steps:

STAT>TESTS>T-test>Stats

We input the value \mu_o =3, \bar X= 2.8, s_x = 2.6, n=1310 and for the alternative we select < \mu_o. Then press Calculate.

And we got the same results.  

3 0
2 years ago
Quinn used compensation to find the product of 37x4. First, she found 40x4=160. Then she adjusted that product by adding 3 group
torisob [31]
<span>Quin is adding the compensation needed instead of subtracting. First she calculated 40x4. She is trying to find 37x4. To do this, she would have to instead gather all the data: 3x4 = 12 40x4 = 160 Subtract instead of add: 40x4 - 3x4 = ? 160 - 12 = 148 The answer is 148.</span>
7 0
3 years ago
Select the correct answer. consider this absolute value function. if function f is written as a piecewise function, which piece
Elanso [62]

x + 3, x > -3 is the piece which will be included in the given function          f(x) = [ x + 3 ]. This can be obtained by removing modulus and finding the value for x.

<h3>Which piece will the function include?</h3>

Given that,  

f(x) = [ x + 3 ] which is an absolute value function.

Therefore,

| x+3 | > 0

By removing the modulus,

x+3 > 0

x > - 3

Hence x + 3, x > -3 is the piece which will be included in the given function          f(x) = [ x + 3 ].

Learn more about absolute value function here:

brainly.com/question/10538556

#SPJ4

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Consider this absolute value function.

f(x) = [ x + 3 ]

If function f is written as a piecewise function, which piece will it include?

A. x + 3, x > 3

B. x + 3, x > -3

C. -x + 3, x < -3

D. -x - 3, x < 3

7 0
1 year ago
Solve the equation 3x+5y=15 for y
larisa86 [58]
<span> 3x+5y = 15
       5y = -3x + 15
         y = -3/5(x) + 3

hope it helps


</span>
7 0
3 years ago
Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x
djverab [1.8K]

I'll assume the ODE is

y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}

Solve the homogeneous ODE,

y'' - 3y' + 2y = 0

The characteristic equation

r^2 - 3r + 2 = (r - 1) (r - 2) = 0

has roots at r=1 and r=2. Then the characteristic solution is

y = C_1 e^x + C_2 e^{2x}

For nonhomogeneous ODE (1),

y'' - 3y' + 2y = e^x

consider the ansatz particular solution

y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x

Substituting this into (1) gives

a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1

For the nonhomogeneous ODE (2),

y'' - 3y' + 2y = e^{2x}

take the ansatz

y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}

Substitute (2) into the ODE to get

b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1

Lastly, for the nonhomogeneous ODE (3)

y'' - 3y' + 2y = e^{-x}

take the ansatz

y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}

and solve for c.

ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16

Then the general solution to the ODE is

\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}

6 0
1 year ago
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