during a particular period a university's information technology office received 20 service orders for problems with printers, of which 8 were laser printers and the rest are inkjet printers a sample of 5 of these service orders is to be selected for inclusion in a customer satisfaction survey what is the probability that 3 were laser printers

0.2384 = 23.84% probability that 3 were laser printers.

Step-by-step explanation:

The printers will be chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

is the number of different combinations of x objects from a set of n elements, given by the following formula.

In this question:

20 orders means that

8 were laser printers, which means that

Sample of 5 means that

What is the probability that 3 were laser printers?

This is P(X = 3).

0.2384 = 23.84% probability that 3 were laser printers.

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step-by-step explanation:

ewd

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.