Answer:
46.6 mph
Step-by-step explanation:
1 Kph is .6213 mph
Answer:
-28 + -1 = -29
-14 + -2 = -16
-7 + -4 = -11
-4 + -7 = -11
-2 + -14 = -16
-1 + -28 = -29
1 + 28 = 29
2 + 14 = 16
4 + 7 = 11
7 + 4 = 11
14 + 2 = 16
28 + 1 = 29
Step-by-step explanation:
Answer:
The weighted average is of 69.94.
Step-by-step explanation:
Weighted average:
The weighed average is found multiplying each grade by its respective weight.
The grades, and weights are:
67 on the lab, with a weight of 23% = 0.23
69 on the first major test, with a weight is 21.5% = 0.215
85 on the second major test, with a weight is 21.5% = 0.215.
63 on the final exam, with a weight of 34% = 0.34.
Weighted average:

The weighted average is of 69.94.
You need to find the least common denominator first which is 4. 3 3/4 can stay the same but you must have like denominators to add so change 1 1/2 to 1 2/4 by multiplying by 2/2. You know have like denominators and you can add whole numbers first 3+1=4 and then the fractions 3/4+2/4= 5/4 which is also 1 1/4. Now you add 4 so your answer is 5 1/4
Answer:
a) 
b) 
c) 
With a frequency of 4
d) 
<u>e)</u>
And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:

Replacing we got:

Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

Part c
The mode is the most repeated value in the sample and for this case is:

With a frequency of 4
Part d
The midrange for this case is defined as:

Part e
For this case we can calculate the deviation given by:

And replacing we got:

And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case