Answer:
x = 52 i hope this helps! :)
Step-by-step explanation:
so first you gotta realize that every triangle has an interior angle measurement sum of 180 degrees
so we are given 3 angles 74 degrees x + 2 and x
we now are going to put this into an equation 74 + x + 2 + x = 180
now lets combine like terms to get 76 + 2x = 180
then we are going to subtract the 76 from both sides 2x = 104
now lets divide both sides by 2 x = 52
so x = 52
we can plug in the value of x into both of the other equations to find the angle measurement
52 degrees
52 + 2 = 54 degrees
and then we have 74 degrees
add the 52, 54, and the 74 to make sure it is equal to 180 degrees
106 + 74 = 180
so yes, x does indeed equal 52 degrees
10 would be 50% so 1 is 5% 5x8 =40
Her experimental probability is 40%
Answer:
Step-by-step explanation:
circumference=πd=π×18=18π in
distance covered in 1 sec=18π×10=180π in.
distance covered in 1 min=180π×60=10800π in
distance covered in 60 min=10800π×60=648000π in=648000π÷12=54000π ft
=54000π÷3=18000π yards=18000π÷1760≈32.13 miles per hour.
Let's add up the given scores to get: 97+38+88+75+83 = 381. Then we divide by 5 since there are five scores here. The mean we get is 381/5 = 76.2
So that explains the "76.2" mentioned.
Now we'll replace the "38" with "58" and repeat the same steps as above.
Add: 97+58+88+75+83 = 401
Divide by five: 401/5 = 80.2
The mean is now 80.2; it has increased compared to the previous mean. We can see that if we move any outliers closer to the main cluster, then the mean will be centered more around the main cluster. In the other direction, if we move the outlier further away from the cluster, then the outlier pulls on the mean. Think of a magnetic or gravitational pull.
The smaller outlier 38 made the mean of 76.2 to be smaller than it should be if we focused on the other values only. In short, if your data set has outliers, then the mean may likely not reflect the true center average value of the group. At that point, the median may be the next best thing or you could use a trimmed mean.
