The pH measurements of water specimens from various locations along a given river basin are Normally distributed with mean 8 and standard deviation 0.3. What is, approximately, the probability that the pH measurement of a randomly selected water specimen is a value between 7.5 and 8.2? A. 0.3003
B. 0.6997
C. 0.9522
D. 0.9902

0.699733

Step-by-step explanation:

Given that :

Mean (m) = 8

Standard deviation (s) = 0.3

Probability that pH is between 7. 5 and 8.2

Probability of pH < 7.5

P(x < 7.5)

The standardized score (Z) :

Z = (x - m) / s

Z = (7.5 - 8) / 0.3 = - 1.6667

p(Z < - 1.6667) = 0.047787 ( Z probability calculator)

Probability of pH < 8.2

P(x < 8.2)

The standardized score (Z) :

Z = (x - m) / s

Z = (8.2 - 8) / 0.3 = 0.6667

p(Z < 0.6667) = 0.74752 ( Z probability calculator)

Hence,

p(Z < 0.6667) - p(Z < - 1.6667)

0.74752 - 0.047787 = 0.699733

the amount you'd have after the work is $1.55

step-by-step explanation:

you subtract 95% from $31.

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1