Suppose approximately 80% of all marketing personnel are extroverts, whereas about 55% of all computer programmers are introverts. (Round your answers to three decimal places.) (a) At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts?

The probability that 10 or more are extroverts is

Step-by-step explanation:

We are given that approximately 80% of all marketing personnel are extroverts, whereas about 55% of all computer programmers are introverts.

Also, a sample of 15 marketing personnel is chosen.

The above situation can be represented through Binomial distribution;

where, n = number of trials (samples) taken = 15 marketing personnel

             r = number of success = 10 or more

            p = probability of success which in our question is % of marketing

                  personnel that are extroverts, i.e; 80%

LET X = Number of marketing personnel that are extroverts

So, it means X ~

Now, Probability that 10 or more are extroverts is given by = P(X 10)

P(X 10)  = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

=

=

= 0.1032 + 0.1876 + 0.2501 + 0.2309 + 0.1319 + 0.0352 = 0.9389

So, the probability that 10 or more are extroverts is 0.9389.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

i think the answer is fried chicken ; >

step-by-step explanation: