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Thepotemich [5.8K]
2 years ago
7

Which sign makes the statement true? |6|? -8 greater than less than or equal to?​

Mathematics
2 answers:
lora16 [44]2 years ago
7 0

Answer: greater than

Step-by-step explanation: Absolute value shows a number's distance from zero on a number line. 6's distance from 0 is still 6. 6 is greater than -8, therefore l6l>-8

kicyunya [14]2 years ago
3 0

Answer:

|6| > -8

Step-by-step explanation:

The absolute value of |6| is 6. Because positives are greater than negatives, 6 is greater than -8.

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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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