K/6 = 15/18 Solve the proportion
2 answers:
Answer: 5
Step-by-step explanation: To solve for <em>k</em> in the given proportion, we use the means-extremes property. The product of the extremes, (k)(18), equals the product of the means, (6)(15).
So we have (k)(18) or 18k = (6)(15) or 90.
Now we have the equation 18k = 90.
Dividing both sides by 18, <em>k = 5</em>.
It's a good idea to check the answer in these types of problems so if we plug a 5 back in for <em>k</em> in the original proportion, we have which is a true statement since 15/18 reduces to 5/6.
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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.