Answer
(a) $\frac{1}{80}$ minute
(b) $80$
(c) The function can be graphed as shown in the figure.
(d) 140/90 higher than the normal of 120/80 mmHg.
Work Step by Step
(a) Compare with a standard sine function $y=a\cdot sin(kx)$ with period $T=\frac{2\pi}{k}$,
function $p(t)$ has a period of $T'=\frac{2\pi}{160\pi}=\frac{1}{80}$ minute.
(b) Since frequency is the reciprocal of the period, the number of heartbeats per minute is given by $\frac{1}{T'}=80$
(c) The function can be graphed as shown in the figure.
(d) Based on the graph, the blood pressure range from 90 to 140 mmHg (140/90)
which are higher than the normal of 120/80 mmHg.