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dolphi86 [110]
2 years ago
7

Jordan uses a linear equation to model the data in a table. Which reasoning might Jordan have used in the decision to model the

data with a linear equation?
The products of corresponding x- and y-values are equal.
The ratios of consecutive y-values are equal for evenly spaced x-values.
The change in y-values is not constant, but the change is constant for evenly spaced x-values.
The change in y-values is constant for evenly spaced x-values.
Mathematics
2 answers:
Anton [14]2 years ago
5 0

ANSWER:

the answer is D.

Mariulka [41]2 years ago
3 0
The last option:The change in y-values is constant for evenly spaced x-values.

Remember that the change in y divided by the change in x is the slope of a line and it is constant.



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Step-by-step explanation:

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80

Step-by-step explanation:

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Question 6
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Answer:

The input for the method is a continuous function f, an interval [a, b], and the function values f(a) and f(b). The function values are of opposite sign (there is at least one zero crossing within the interval). Each iteration performs these steps: Calculate c, the midpoint of the interval, c = a + b2.

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2 years ago
Zappos is an online retailer based in Nevada and employs 1,300 employees. One of their competitors, Amazon, would like to test t
Amanda [17]

Answer:

t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402    

df = n-1= 22-1=21

t_{\alpha/2}= -2.08

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

Step-by-step explanation:

Data given

\bar X=33.9 represent the sample mean

s=4.1 represent the sample standard deviation

n=22 sample size  

\mu_o =26 represent the value that we want to test

\alpha=0.025 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)  

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 36 years old, the system of hypothesis would be:  

Null hypothesis:\mu \geq 36  

Alternative hypothesis:\mu < 36  

The statistic is given by:

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

And replacing we got:

t=\frac{33.9-36}{\frac{4.1}{\sqrt{22}}}=-2.402    

Now we can calculate the critical value but first we need to find the degreed of freedom:

df = n-1= 22-1=21

So we need to find a critical value in the t distribution with df =21 who accumulates 0.025 of the area in the left and we got:

t_{\alpha/2}= -2.08

Since the calculated values is lower than the critical value we have enough evidence to reject the null hypothesis at the significance level of 2.5% and we can say that the true mean is lower than 36 years old

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3 years ago
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