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Zolol [24]
3 years ago
11

Drag numbers to use a negative factor to factor the expression –5g + 15h – 25. Numbers may be used once, more than once, or not

at all.
–1 –2 –3 –4 –5 1 2 3 4 5
(_____g –_____ h +_____)
Mathematics
1 answer:
Gala2k [10]3 years ago
6 0

Answer:

(5g –15h +25)

hope dis helps ^-^

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A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a populat
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Answer:

t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2      

p_v =P(t_{15}>2)=0.0320  

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.  

Step-by-step explanation:

1) Data given and notation      

\bar X=25 represent the sample mean      

s=2 represent the standard deviation for the sample      

n=16 sample size      

\mu_o =24 represent the value that we want to test    

\alpha 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 (variable of interest)  

Confidence =0.95 or 95%

\alpha=0.05

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the mean is higher than 24, the system of hypothesis would be:      

Null hypothesis:\mu \leq 24      

Alternative hypothesis:\mu > 24      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic      

We can replace in formula (1) the info given like this:      

t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

df=n-1=16-1=15  

Since is a one-side upper test the p value would be:      

p_v =P(t_{15}>2)=0.0320  

Conclusion      

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.      

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