<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>

4 0

3 years ago

The annual 2-mile fun-run is a traditional fund-raising event to support local arts and sciences activities. It is known that th

victus00 [196]

Answer: 0.0228

Step-by-step explanation:

Given : The mean and the standard deviation of finish times (in minutes) for this event are respectively as :-

$\mu=30\\\\\sigma=5.5$

If the distribution of finish times is approximately bell-shaped and symmetric, then it must be normally distributed.

Let X be the random variable that represents the finish times for this event.

z score : $z=\dfrac{x-\mu}{\sigma}$

$z=\dfrac{19-30}{5.5}=-2$

Now, the probability of runners who finish in under 19 minutes by using standard normal distribution table :-

$P(X$

Hence, the approximate proportion of runners who finish in under 19 minutes = 0.0228

8 0

2 years ago