Answer:
y=32
Step-by-step explanation:
The relationship between the value of x and y is multiplying by 8, therefore when x equals 4, y equals 32.
A cheerful teen willing to help,
listening to "Rise Up,"
stay salty...
Answer:
Minimum cycle time = time for longest task = 2.4 minutes
Maximum cycle time = sum of the times to complete all the tasks = 18 minutes
Step-by-step explanation:
For the minimum cycle time, the reasoning is that all the tasks can be done simultaneously, at some during the assembling task. Hence, minimum time to complete one cycle is the time it takes to complete task that takes the longest time to complete.
For the maximum cycle time, the reasoning here is that each task is done after the completion of the previous task. The total time taken to finish the task will then be the sum of the time taken to complete each task. This usually is at the beginning of operations.
Answer:
This cannot be solved but it can be simplified
first we open the bracket
so we have
a²- abx/x = 2/3
next we cross multiply so it will be 3(a² - abx) = 2x
we get
3a² - 3abx = 2x
Next we factorise so we can get
3a(a-bx) =2x
I dont know what exactly you where asked but I hop this helps
The initial fee of $50 is essentially the y intercept because this is the value when x = 0 (x is the number of labor-hours). So b = 50.
The slope is m = 30 because each increase of 1 hour leads to the cost bumping up by 30 dollars. In other words, slope = rise/run = (change in cost)/(change in hours) = 30/1
So we plug m = 30 and b = 50 into the y = mx+b formula to get y = 30x+50
Replace y with f(x) to get f(x) = 30x+50
The linear function for the cost is f(x) = 30x+50
Note: Some books may use other letters (instead of x and f(x)), but the idea is still the same
Once you know the cost function, you replace x with 4.5 to find the amount it will cost to have a painter work for 4.5 hours.
f(x) = 30x+50
f(4.5) = 30*4.5+50
f(4.5) = 135+50
f(4.5) = 185
It will cost 185 dollars to have the painter work for 4.5 hours
Answer:
a) 30 kangaroos in 2030
b) decreasing 8% per year
c) large t results in fractional kangaroos: P(100) ≈ 1/55 kangaroo
Step-by-step explanation:
We assume your equation is supposed to be ...
P(t) = 76(0.92^t)
__
a) P(10) = 76(0.92^10) = 76(0.4344) = 30.01 ≈ 30
In the year 2030, the population of kangaroos in the province is modeled to be 30.
__
b) The population is decreasing. The base 0.92 of the exponent t is the cause. The population is changing by 0.92 -1 = -0.08 = -8% each year.
The population is decreasing by 8% each year.
__
c) The model loses its value once the population drops below 1/2 kangaroo. For large values of t, it predicts only fractional kangaroos, hence is not realistic.
P(100) = 75(0.92^100) = 76(0.0002392)
P(100) ≈ 0.0182, about 1/55th of a kangaroo
