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jolli1 [7]
2 years ago
5

Can anyone help me with dis

Mathematics
1 answer:
dolphi86 [110]2 years ago
7 0

Answer:

Slope is 2.

Step-by-step explanation:

So to find slope, we need to take the change in the y value, and the change in the x value, and then divide them. This may be confusing, so here is just 3 formulas in 3 seperate steps that we use to find it:

Step 1 - x_2-x_1=change_.in_.x

Step 2 - y_2-y_1=Change_.in_.y

Step 3 - \frac{Change_. in_. y}{Change_.in_.x}=slope

So lets start by doing step one.

All we need is to look for two x values. Lets use 5 and 6:

6-5=1

So 1 is our change of x. This is our 1st step.

Next we can use 2 y values, note that these should be the 2 y values that go wiht the x values above. So this means we have to use -3 and -1:

-1-(-3) = 2

So 2 is our change in y. This is our 2nd step.

Now lets plug this into:

\frac{Change_. in_. y}{Change_.in_.x}=slope:

\frac{2}{1}  = 2

So our slope is 2. Completing our 3rd and final step.

Hope this helps!

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Answer:

t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974      

p_v =P(t_{99}>8.974)=9.43x10^{-15}  

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 to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

Step-by-step explanation:

1) Data given and notation      

\bar X=23500 represent the sample mean  

s=3900 represent the sample standard deviation      

n=100 sample size      

\mu_o =2000 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)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the true mean is higher than 20000, the system of hypothesis would be:      

Null hypothesis:\mu \leq 2000      

Alternative hypothesis:\mu > 2000      

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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{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974      

Calculate the P-value      

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

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

p_v =P(t_{99}>8.974)=9.43x10^{-15}  

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 to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

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