The pH of randomly selected swimming pools follows a normal distribution with mean 7.4 and standard deviation 0.22. A pool should have a pH between 7.2 and 7.8 so that chlorine treatment is effective. What percentage of pools having an acceptable pH level

78.383%

Step-by-step explanation:

The pH of randomly selected swimming pools follows a normal distribution with mean 7.4 and standard deviation 0.22. A pool should have a pH between 7.2 and 7.8 so that chlorine treatment is effective.

Using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

For x = 7.2

z = 7.2 - 7.4/0.22

z = -0.90909

Probability value from Z-Table:

P(x = 7.2) = 0.18165

For x = 7.8

z = 7.8 - 7.4/0.22

z = 1.81818

Probability alue from Z-Table:

P(x = 7.8) = 0.96548

The probability of pools having an acceptable pH level(7.2 and 7.8)

= P(x = 7.8) - P(x = 7.2)

= 0.96548 - 0.18165

= 0.78383

Converting to percentage

0.78383 × 100

= 78.383%

The percentage of pools having an acceptable pH level is 78.383%

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.

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐