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Debora [2.8K]
3 years ago
6

2/5ab What are the factors of the expression? Drag all of the factors of the expression into the box. Factors −1 2 5 2 5 a b

Mathematics
1 answer:
ELEN [110]3 years ago
6 0
2/5, a, and b is the answer
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A study is conducted to determine if a newly designed text book is more helpful to learning the material than the old edition. T
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Answer:

We need to conduct a hypothesis in order to check if the mean is higher than 75, the system of hypothesis are :  

Null hypothesis:\mu \leq 75  

Alternative hypothesis:\mu > 75  

A. One-Tailed

z=\frac{82.2-75}{\frac{10.4}{\sqrt{10}}}=2.189  

p_v =P(z>2.189)=0.0143  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Step-by-step explanation:

Data given and notation  

We have the following data: 84 94 80 88 77 68 90 74 96 71

We can calculate the sample mean with the following formula:

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

And replacing we got:

\bar X=82.2 represent the sample mean  

\sigma=10.4 represent the population standard deviation  

n=10 sample size  

\mu_o =75 represent the value that we want to test  

\alpha=0.01 represent the significance level for the hypothesis test.  

z 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 mean is higher than 75, the system of hypothesis are :  

Null hypothesis:\mu \leq 75  

Alternative hypothesis:\mu > 75  

A. One-Tailed

Since we know the population deviation, 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{82.2-75}{\frac{10.4}{\sqrt{10}}}=2.189  

P-value  

Since is a ONE-TAILED  test the p value would given by:  

p_v =P(z>2.189)=0.0143  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

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A car travels 35 mph for r+2 hours,how far does it travel
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Recall d = rt, distance = rate * time.

\bf d=rt\qquad 
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r=\stackrel{mph}{35}\\
t=\stackrel{hours}{r+2}
\end{cases}\implies d=35(r+2)\implies d=35r+70
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