Suppose the probability that a randomly selected man, aged 55-59, will die of cancer during the course of the year is startfraction 300 over 100 comma 000 endfraction . how would you find the probability that at least 1 man out of 1,000 of this age will die of cancer during the course of the year?

The question is an example of binomial distribution.

P(x=r) = nCr* p^r* q^(n-r) ( nCr = n! / ( (n-r)!*r! )

p = probability of die = 300/100000 = .003

P ( at least one man would die) = 1-P(at most one man would die)

= 1 - [ P(X =0) + P(X= 1)

= 1- [ {1000C0(.003)^0 *(.997)^1000} + { 1000C1(.003)^1 *(.997)^999} ]

= .801

y-1700=200(x+4)

step-by-step explanation:

200(x+4) is 200x + 800 she is saving 200 per week x is the weeks and the -1700 is. us that’s howmuch she already had

5: 20

step-by-step explanation: