Answer:
Let the smaller integer be x and greater integer be (x+1)
x+(x+1)=91
2x+1=91
2x=91-1=90
x=90/2
x=45
Answer:
Step-by-step explanation:
In March, Your Co. will collect 20% of January's sales, 30% of February's sales, and 50% of March's sales:
.20×50 +.30×40 +.50×60 = 10 +12 +30 = 52
Similarly, in April, collections will be ...
.20×40 + .30×60 + .50×30 = 8 +18 +15 = 41
See the attached picture:
C. add 3 to 15 because the sequence is +3 every time
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)
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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.
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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.
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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