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alekssr [168]
2 years ago
14

Calculate the t-value given the following: hypothesis test on population mean;

Mathematics
1 answer:
11Alexandr11 [23.1K]2 years ago
6 0

The  t-value given the following: hypothesis test on population mean is -2.

<h3>What is T- value ?</h3>

T-value in statistics can be defined as the ratio of the difference between the mean of the two sample data and the variation in the sample data

It is given  by

\rm t = \dfrac{x- \mu}{\dfrac{s}{\sqrt{n}}}

Here in the question all the values are given

x= 33; μ = 35; s = 5; n = 25.

Substituting the values to determine t

\rm t = \dfrac{33- 35}{\dfrac{5}{\sqrt{25}}}

\rm t = \dfrac{-2}{\dfrac{5}{5}}\\\\\\t = -2

Therefore the t-value given the following: hypothesis test on population mean is -2.

To know more about t value

brainly.com/question/14859752

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2x - 5y = 4<br> 3x - 7y = 5
VMariaS [17]

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(-3,-2)

Step-by-step explanation:

2x-5y=4

3x-7y=5

solve the equation

x=2+5/2y

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substitute the value of x into an equation

3(2+5/2y)-7y=5

multiply 3 to 2. and. 3 to 5/2y

6+15/2y-7y=5

minus 6 to 5 and. add 15/2y to -7y

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divide both sides by -1/2

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substitute the value of -2 into an equation

x=2+5/2•(-2)

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add but since 5 is negative then subtract the numbers

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3 0
3 years ago
The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section
lisabon 2012 [21]

Answer:

\therefore y_2(x)=-\frac{e^{-6x}}{8}

The general solution is

y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}

Step-by-step explanation:

Given differential equation is

y''-4y'+4y=0

and y_1(x)=e^{2x}

To find the y_2(x) we are applying the following formula,

y_2(x)=y_1(x)\int \frac{e^{-\int P(x) dx}}{y_1^2(x)} \ dx

The general form of equation is

y''+P(x)y'+Q(x)y=0

Comparing the general form of the differential equation to the given differential equation,

So, P(x)= - 4

\therefore y_2(x)=e^{2x}\int \frac{e^{-\int 4dx}}{(e^{2x})^2}dx

           =e^{2x}\int \frac{e^{-4x}}{e^{4x}}dx

           =e^{2x}\int e^{-4x-4x} \ dx

            =e^{2x}\int e^{-8x} \ dx

            =e^{2x}. \frac{e^{-8x}}{-8}

           =-\frac{e^{-6x}}{8}

\therefore y_2(x)=-\frac{e^{-6x}}{8}

The general solution is

y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}

4 0
3 years ago
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