0.57619047 hoped this helped
Answer:
Distance between Montpelier and Columbia is 1020
Step-by-step explanation:
51 mi/hr average:
d = 51 t
60 mi/hr average:
d t = 60*(t-3)
Substitute (1) into (2)
51 t = 60*(t-3)
51 t = 60 t + 180
9 t = 180
t = 20
d = 51 t
= 51 (20)
= 1020
let see distance is 1020 lets check
51 t = 60 * (t-3)
51 (20)= 60* (20-3)
1020 = 60*17
1020 = 1020
Based on the data set, the variance for this sample of tests is 4
<h3>How to find the variance for this sample of tests?</h3>
The sample of test is given as:
78, 79, 79, 81, and 83.
Calculate the mean, using:
Mean = Sum/Count
So, we have
Mean = (78 + 79 + 79 + 81 + 83)/5
Evaluate
Mean = 80
The variance is calculated as

This gives
= [(78 - 80)^2 + (79 - 80)^2 + (79- 80)^2 + (81- 80)^2 + (83- 80)^2)]/[5 - 1]
Evaluate the numerator and the denominator.
So, we have:
=16/4
Evaluate the quotient
= 4
Hence, the variance for this sample of tests is 4
Read more about variance at:
brainly.com/question/15858152
#SPJ1
Answer:69 2/9 - 31 1/9 - (- 12 4/9) =
69 2/9 - 31 1/9 + 12 4/9 =
81 6/9 - 31 1/9 =
50 5/9 or 455/9 or 50.56 <==
Step-by-step explanation: