Aplumber loads his truck each morning with faucets that will be needed for the service calls and other emergency calls that come in that day. based on past experience, the number of faucets required each day (n) has the following distribution: p(0)equals.05; p(1)equals.25; p(2)equals.5; p(3)equals.15; p(4)equals.05. the expected value of n is then 1.9 faucets required per day. each faucet requires 25 minutes to install. what is the expected number of minutes the plumber will spend each day installing faucets?

47.5 minutes.

Step-by-step explanation:

Data provided:

p(0) = 0.05

p(1) = 0.25

p(2) = 0.5

p(3) = 0.15

p(4) = 0.05

Time required by each faucet to install = 25 minutes

Therefore,

The expected number of minutes spent by the plumber each day installing faucets

= ∑ ( xp(x) × time )

= ( 0 × 0.05 × 25 ) + ( 1 × 0.25 × 25 ) + ( 2 × 0.5 × 25 ) + ( 3 × 0.15 × 25 ) + ( 4 × 0.05 × 25 )

= 6.25 + 25 + 11.25 + 5

= 47.5 minutes.

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

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.