PLEASE help!! Given the table below, determine if the data represents a linear or an exponential function and find a possible fo
2 answers:
x 0 1 2 3 4
f(x) 12.5 13.75 15.125 16.638 18.301
If the change in x values and change in y values are constant then the data will be linear.
So, change in x from 0 to 1 is 1-0=1
Similarly change in x from 1 to 2 is 2-1=1
So, change in x's is constant in the given data.
Now let's check for change in y values.
Change in y from 12.5 to 13.75 is 13.75-12.5=1.25
Similarly change in y from 13.75 to 15.125 is 15.125-13.75=1.375
Changes in y's are not constant.
So, the given data is not linear.
Let's assume the exponential function is:
Let's take any two points from the data and plug into the above equation to get the equation. Let's plug in the first point (0, 12.5) in the above function. So,
Since
So, a=12.5
Next step is to plug in the second point (1, 13.75) and a=12.5 in the same equation to get the value of b. Hence,
13.75=12.5b
Dividing each sides by 12.5.
So, b=1.1
Now we can plug in the value of a and b to get the exponential formula. Hence,
So, the correct choice is a.
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Answer:
Step-by-step explanation:
Hello!
The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.
The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.
You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:
Sample 1 - Machine 1
X₁: Net volume of a plastic bottle filled by the machine 1.
n₁= 10
X[bar]₁= 16.02
S₁= 0.03
Sample 2 - Machine 2
X₂: Net volume of a plastic bottle filled by the machine 2
n₂= 10
X[bar]₂= 16.01
S₂= 0.03
With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:
X₁~N(μ₁;δ₁²)
X₂~N(μ₂;δ₂²)
Both population variances are unknown you have to conduct a homogeneity of variances test to see if they are equal (if they are you can conduct a pooled t-test) or different (if the variances are different you have to use the Welche's t-test)
H₀: δ₁²/δ₂²=1
H₁: δ₁²/δ₂²≠1
α: 0.05
Using a statistic software I've calculated the test
p-value 0.6168
The p-value is greater than the significance level, so the decision is to not reject the null hypothesis then you can conclude that both population variances for the net volume filled in the plastic bottles by machines 1 and 2 are equal.
To study the difference between the population means you can use
The hypotheses of interest are:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
α: 0.05
The p-value for this test is: 0.882433
The p-value is greater than the significance level, the decision is to not reject the null hypothesis.
Using a significance level of 5%, there is no significant evidence to reject the null hypothesis. You can conclude that the population average of the net volume of the plastic bottles filled by machine one and by machine 2 are equal.
I hope this helps!
