Determine whether the following value is a continuous random variable, discrete random variable, or not a random variable. a. The exact time it takes to evaluate 67 plus 29 b. The usual mode of transportation of people in City Upper A c. The height of a randomly selected person d. The number of people in a restaurant that has a capacity of 100 e. The time it takes for a light bulb to burn out f. The number of statistics students now doing their homework

a. Continuous random variable.

b. Not a random variable.

c. Continuous random variable.

d. Discrete random variable.

e. Continuous random variable.

f. Discrete random variable.

Step-by-step explanation:

A random variable is a numerical quantity resulted by a random experiment.

Continuous random variable contains measurable quantities while discrete random variable contains countable quantities.

a.

The exact time it takes to evaluate 67 plus 29 is a continuous random variable because exact time will be measurable as it can contains minutes and seconds etc. If the number of minutes are concerned to evaluate 67 plus 29 then it will be a discrete variable because number of minutes can be counted like 1 minutes, 2 minutes etc which is not the case here.

b.

The usual mode of transportation of people in City Upper A  is not random variable because it contains non numerical values such as cars,buses and bicycles etc.

c.

The height of a randomly selected person s a continuous random variable because height will be measurable.

d.

The number of people in a restaurant that has a capacity of 100 is a discrete variable because number of people in a restaurant can be counted like 54 people, 75 people etc.

e.

The time it takes for a light bulb to burn out is a continuous random variable because  time will be measurable as it can contains hours and minutes etc.

f.

The number of statistics students now doing their homework is a discrete variable because number of students doing their homework can be counted like 4 students, 6 students etc.

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.

his chances of getting a coin of higher value is 4/19

step-by-step explanation: