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netineya [11]
2 years ago
6

I need this answered please

Mathematics
1 answer:
TiliK225 [7]2 years ago
4 0
The answers are

1: False
2: True
3: True
4: False
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-4/5 = -16/20, so the new expression is (-16/20)+(3/20)
then, -16 + 3 is -13, so the solution is -13/20
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Use (<, >, =) to compare 2⁄3 and 3⁄4. A. None of the above B. 2⁄3 > 3⁄4 C. 2⁄3 < 3⁄4 D. 2⁄3 = 3⁄4
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A researcher tests five individuals who have seen paid political ads about a particular issue. These individuals take a multiple
devlian [24]

Answer:

t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606    

The degrees of freedom are given by:

df=n-1=5-1=4  

The p value wuld be given by:

p_v =2*P(t_{(4)}>2.606)=0.060  

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

Step-by-step explanation:

Information given

48, 41, 40, 51, and 50

The sample mean and deviation can be calculated with these formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=46 represent the mean height for the sample  

s=5.148 represent the sample standard deviation

n=5 sample size  

\mu_o =40 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test

Hypothesis to test

We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:  

Null hypothesis:\mu = 40  

Alternative hypothesis:\mu \neq 40  

The statistic is given by:

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

Replacing we got:

t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606    

The degrees of freedom are given by:

df=n-1=5-1=4  

The p value wuld be given by:

p_v =2*P(t_{(4)}>2.606)=0.060  

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

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11 2/3 seconds at a speed of 12 m/s
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