SCSA Mathematics Applications Growth and decay in sequences

$8700 - 8400 = 300$

Therefore the constant rate is 300 cells/$\mu$L.

$T_n = 8.4 + (n - 1)(0.3)$
$T_n = 8.1 + 0.3n$

$T_{12} = 8.1 + 0.3(12)$
$T_{12} = 11.7$

Therefore 11 700 cells/$\mu$L.

$T_1 = (-0.5)(20) + 20 = 10$
$T_2 = (-0.5)(10) + 20 = 15$

Therefore the white blood cell count in the body is 15 000 cells/$\mu$L at the end of the second hour.

$x = -0.5x + 20 \Rightarrow x = \frac{20}{1.5} = \frac{200}{15} = \frac{40}{3} = 13\frac{1}{3} \approx 13.3$

i.e. approximately 13 300 cells/$\mu$L, which is greater than 13 000

Therefore the given medication will ensure the person's white blood cell count does not fall below 13 000 cells per microlitre.

$T_{n+1} = T_n + 3, \quad T_1 = 250$

$250 + 253 + 256 + 259 = 1018$ kilometres

The 220 represents the distance travelled in kilometres by Simon by the end of day one.
1.06 represents a 6% increase in distance travelled each day.

Therefore Simon travels 330.8 km on day 8 which is Saturday, 9 January.

Josh $\rightarrow$ total distance after 10 days = 2635 kilometres
total distance after 11 days = 2915 kilometres
Therefore Josh will arrive on the $11^{\text{th}}$ day.

Simon $\rightarrow$ total distance after 9 days = 2528.09 kilometres
total distance after 10 days = 2899.77 kilometres
Therefore Simon will arrive on the $10^{\text{th}}$ day.

Simon will arrive in Alice springs first.

$309 = a(380) + b$ ...(1)
$269.95 = a(309) + b$ ...(2)

$a = 0.55, \; b = 100$

$T_{n+1} = T_n + 50, T_1 = 4000$

$T_n = 4000 + (n-1)(50)$
$T_n = 3950 + 50n$

$T_{12} = 3950 + 50(12) = 4550$

Therefore, the monthly wage for December 2020 is ＄4550

$12 \times 4 = 48$
$T_{48} = 3950 + 50(48) = 6350$

Therefore, the monthly wage for March 2024 is $6350 + 60 + 60 + 60 = ＄6530$