Answer:
• Because the sets are not symmetrical, the IQR should be used to compare the data sets.
• Because the sets contain outliers, the median should be used to compare the data sets.
• The mean and mode cannot be accurately determined based on the type of data display.
Step-by-step explanation:
When we observe the set of given data above, we can denote that the data obtained by comparing the height of students from class 1 and class 2 would not be similar hence we can say this obtained data is not symmetrical.
Due to the fact that this data is is obtained from different classes it is certain that there would be variations in the data when measuring the heights of the students and an error may occurs. These variations are referred to as OUTLIERS.
Therefore, Median or Interquartile range is the appropriate measure to be used for comparing the data sets.
We can create a parabola equation of the trajectory using
the vertex form:
y = a (x – h)^2 + k
The center is at h and k, where h and k are the points at
the maximum height so:
h = 250
k = 120
Therefore:
y = a (x – 250)^2
+ 120
At the initial point, x = 0, y = 0, so we can solve for
a:
0 = a (0 – 250)^2 + 120
0 = a (62,500) + 120
a = -0.00192
So the whole equation is:
y = -0.00192 (x – 250)^2 + 120
So find for y when the golf ball is above the tree, x =
400:
y = -0.00192 (400 - 250)^2 + 120
y = 76.8 ft
So the ball cleared the tree by:
76.8 ft – 60 ft = 16.8 ft
Answer:
16.8 ft
Answer:
$95.78
Step-by-step explanation:
f(t) = 300t / (2t² + 8)
t = 0 corresponds to the beginning of August. t = 1 corresponds to the end of August. t = 2 corresponds to the end of September. So on and so forth. So the second semester is from t = 5 to t = 10.
$T₂ = ∫₅¹⁰ 300t / (2t² + 8) dt
$T₂ = ∫₅¹⁰ 150t / (t² + 4) dt
$T₂ = 75 ∫₅¹⁰ 2t / (t² + 4) dt
$T₂ = 75 ln(t² + 4) |₅¹⁰
$T₂ = 75 ln(104) − 75 ln(29)
$T₂ ≈ 95.78
