Answer:
10
Step-by-step explanation:
<span>If 100C2 has a value of 4,950, what is the value of 100C98?
10C2 is the number of groups of 2 that can be formed from 100 objects.
Note: Each time a group of 2 is selected you automatically form
a second group of 98. Therefore: 10C2 = 10C98 = 4950 is your Answer</span>
1 = 1, 2, 3, 4, 5
4= 4, 8, 12, 16, 20
5= 5, 10, 15, 20, 25
Answer:
The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h
Step-by-step explanation:
At noon the location of Lan = 300 km north of Makenna
Lan's direction = South
Lan's speed = 60 km/h
Makenna's direction and speed = West at 75 km/h
The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km
The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km
The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km
Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.
By Pythagoras' theorem, we have;
s² = x² + y²
The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61
ds²/dt = dx²/dt + dy²/dt
2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt
2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900
ds/dt = 900/(2×30·√61) ≈ 1.92
The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h
Answer:
Test statistic = - 0.851063
- 2.520463
Step-by-step explanation:
H0 : μ ≥ 15
H1 : μ < 15
Sample mean, xbar = 14.5
Sample standard deviation, s = 4.7
Sample size = 64
Teat statistic :
(xbar - μ) ÷ (s/√(n))
(14.5 - 15) ÷ (4.7/√(64))
= - 0.851063
The critical value at α = 0.05
Using the T - distribution :
Degree of freedom, df = 64 - 1 = 63
Tcritical(0.05, 63) = 1.6694
Test statistic - critical value
-0.851063 - 1.6694
= - 2.520463
