If you double the dimensions of the tank, you also double the volume of the tank.
Therefore, volume of the new touch tank = 2000 x 2 = 4000 ft^3
Answer:
56 number of ways
Step-by-step explanation:
This question is a combination question since it involves selection.
Generally, if r objects are to be selected from n pool of objects, this can be done in nCr number of ways.
nCr = n!/(n-r)!r!
If a manager receives 8 applications for a specific position and wants to narrow it down to 5, the number of ways he can do this is 8C5
8C5 = 8!/(8-5)!5!
= 8!/3!5!
= 8*7*6*5!/3*2*5!
= 8*7*6/3*2
= 8*7
= 56 number of ways.
This means that the manager can rank 5 applications in 56 number of ways
Since the sum of scores must be at least 289:
72+78+70+x ≥ 289
Answer:
Step-by-step explanation:
a) For a prime numbers we have array with 2 rectangulars R1: a=1 and b=prime number; R2: a=prime number and b=1. Both has the same are, that prime number.
b) For a composite number which are not square number we have rectanular array with even numbers of ractangulars. For example, number 6.
R1: a=1,b=6; R2: a=2,b=3; R3: a=3, b=2; R4: a=6,b=1. Each rectangular has the same area, 6.
c) The square number we alway have te odd number of rectanulars, because of the square a=x,b=x can not be simetric. For example 16.
R1: a=1,b=16; R2: a=2 , b=8; R3: a=4,b=4; R4: a=8, b=2; R5:a=16,b=1.Each rectangular has the same area, 16.
Answer:
MArginal productivity: 
We can interpret this as he will reduce his time an <em>additional </em>0.0002 seconds for every <em>additional </em>yard he trains.
Step-by-step explanation:
The marginal productivy is the instant rate of change in the result for an increase in one unit of a factor.
In this case, the productivity is the time he last in the 100-yard. The factor is the amount of yards he train per week.
The marginal productivity can be expressed as:

where dt is the variation in time and dL is the variation in training yards.
We can not derive the function because it is not defined, but we can approximate with the last two points given:

Then we can interpret this as he will reduce his time an <em>additional </em>0.0002 seconds for every <em>additional </em>yard he trains.
This is an approximation that is valid in the interval of 60,000 to 70,000 yards of training.