Answer: 18.7 mph
Step-by-step explanation:
We want a unit rate where 1 is in the denominator, so we divide top and bottom by 41.1
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Solve /x-5/=3 A. x=2, x=8 B. x=-8, x=8 C. x=-2, x=2 D. x=-8, x=-2
1 answer:
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For this case we have the following difference equation:
Applying separable variables we have:
Integrating both sides we have:
applying exponential to both sides:
For y (1) = 1 we have:
Thus, the particular solution is:
Whose domain is all real.
Answer:
y = 0.2 * exp ((3/2) x ^ 2)
Domain: all real numbers
Applying separable variables we have:
Integrating both sides we have:
applying exponential to both sides:
For y (1) = 1 we have:
Thus, the particular solution is:
Whose domain is all real.
Answer:
y = 0.2 * exp ((3/2) x ^ 2)
Domain: all real numbers
Hi!
To compare this two sets of data, you need to use a t-student test:
You have the following data:
-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph
-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph
You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.
To calculate the degrees of freedom you need to use the following equation:
≈34
The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10
So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.
To compare this two sets of data, you need to use a t-student test:
You have the following data:
-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph
-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph
You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.
To calculate the degrees of freedom you need to use the following equation:
The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10
So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.