Answer:
Step-by-step explanation:
v = {[(20sin36°)i + (20cos36°)j] + 10i} mi/h
vE = 20sin36º + 10 = 21.76 mi/h
vN = 20cos36° = 16.18 mi/h
v = √(vE2 + vN2) = √(21.762 + 16.182) mi/h = 27.12 mi/h
θ = tan-1(vN/vE) = tan-1(16.18/21.76) = 36.6º north of east
The <em><u>correct answer</u></em> is:
Ken will have run 3 laps and Hamid will have run 4.
Explanation:
To find this, we first find the number of seconds that will have passed when they meet again. We use the LCM, or least common multiple, for this. First we find the prime factorization of each number:
80 = 10(8)
10 = 5(2)
8 = 2(4)
4 = 2(2)
80 = 2(2)(2)(5)(2)
60 = 10(6)
10 = 5(2)
6 = 2(3)
60 = 2(2)(3)(5)
For the LCM, we multiply the common factors by the uncommon. Between the two numbers, the common factors are 2, 2 and 5. This makes the uncommon 2, 2, and 3, and makes our LCM
2(2)(5)(2)(2)(3) = 240
This means every 240 seconds they will both be at the start line.
Since Ken completes a lap in 80 seconds, he completes 240/80 = 3 laps in 240 seconds.
Since Hamid completes a lap in 60 seconds, he completes 240/60 = 4 laps in 240 seconds.
Answer:
No, Tony is not correct. Solving the inequality tells us that x is greater than or equal to 37.5. Since the class must wash a whole number of cars, they need to wash at least 38 cars.
this will help
Answer:
The restocking level is 113 tins.
Step-by-step explanation:
Let the random variable <em>X</em> represents the restocking level.
The average demand during the reorder period and order lead time (13 days) is, <em>μ</em> = 91 tins.
The standard deviation of demand during this same 13- day period is, <em>σ</em> = 17 tins.
The service level that is desired is, 90%.
Compute the <em>z</em>-value for 90% desired service level as follows:
*Use a <em>z</em>-table for the value.
The expression representing the restocking level is:
Compute the restocking level for a 90% desired service level as follows:
Thus, the restocking level is 113 tins.
