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jek_recluse [69]
3 years ago
10

Solve using countdown method 890-460​

Mathematics
1 answer:
serious [3.7K]3 years ago
7 0

Answer:

430

Step-by-step explanation:

89-46=43

=430

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A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and r
dsp73

Answer:

t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236    

Now we can find the degrees of freedom:

df=n-1=80-1=79  

Now we can calculate the p value with the alternative hypothesis using the following probability:

p_v =P(t_{(79)}>2.236)=0.0141  

Using a significance level of 0.01 or 1% we have enough evidence to FAIL to reject the null hypothesis and we can conclude that shoppers participating in the loyalty program NOT spent more on average than typical shopper at this significance level assumed

Step-by-step explanation:

Information given

\bar X=130 represent the sample mean

s=40 represent the sample standard deviation

n=80 sample size  

\mu_o =120 represent the value to verify

t would represent the statistic  

p_v represent the p value

System of hypothesis

We want to verify if  shoppers participating in the loyalty program spent more on average than typical shoppers (120) , the system of hypothesis would be:  

Null hypothesis:\mu \leq 120  

Alternative hypothesis:\mu > 120  

The statistic for this case is given by:

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

Replacing the info given we got:

t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236    

Now we can find the degrees of freedom:

df=n-1=80-1=79  

Now we can calculate the p value with the alternative hypothesis using the following probability:

p_v =P(t_{(79)}>2.236)=0.0141  

Using a significance level of 0.01 or 1% we have enough evidence to FAIL to reject the null hypothesis and we can conclude that shoppers participating in the loyalty program NOT spent more on average than typical shopper at this significance level assumed

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1.995 x 10 to the negative power of 23?
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Domain are the x values

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Your answer would be 3 
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