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aleksandrvk [35]
2 years ago
4

A computer had 2 gigabytes of data stored on it when Jackie bought it, and she is storing an additional 3.5 gigabytes per year,

as shown in the table below. Which of these statements is correct about the linear function that represents this situation? Amount of Data Stored on Jackie’s Computer Time (year) 0 1 2 3 4 Gigabytes of Data 2 2 + 3.5 = 5.5 5.5 + 3.5 = 9 9 + 3.5 = 12.5 12.5 + 3.5 = 16 The initial value represents the 2 gigabytes of data stored on the computer when Jackie bought it, and the rate of change represents the 3.5 gigabytes per year that Jackie is storing. The initial value represents the 2 gigabytes per year that Jackie is storing, and the rate of change represents the 3.5 gigabytes of data stored on the computer when Jackie bought it. The initial value represents the 3.5 gigabytes of data stored on the computer when Jackie bought it, and the rate of change represents the 2 gigabytes per year that Jackie is storing. The initial value represents the 3.5 gigabytes per year that Jackie is storing, and the rate of change represents the 2 gigabytes of data stored on the computer when Jackie bought it. Mark this and return
Mathematics
1 answer:
bekas [8.4K]2 years ago
8 0

Answer:The initial value represents the 2 gigabytes of data stored on the computer when Jackie bought it, and the rate of change represents the 3.5 gigabytes per year that Jackie is storing.

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Answer:

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

p_v =2*P(z>1.413)=0.158  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation  

\bar X=5.0611 represent the sample mean

\sigma=0.2803 represent the population standard deviation for the sample  

n=42 sample size  

\mu_o =5 represent the value that we want to test

\alpha=0.1 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 5, the system of hypothesis would be:  

Null hypothesis:\mu = 5  

Alternative hypothesis:\mu \neq 5  

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-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:  

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

P-value

Since is a two sided test the p value would be:  

p_v =2*P(z>1.413)=0.158  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

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There is many ways to do this but i did it this way.

180-123.75= 56.25

56.25/5=11.25

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