A chemical plant is required to maintain ambient sulfur levels in the working environment atmosphere at an average level of no more than 12.50. The results of 15 randomly timed measurements of the sulfur level produced a sample mean of Y = 14.82 and a sample standard deviation of s = 2.91. Required:
Is there evidence that the chemical plant is in violation of the working code?

Since the calculated value of z= 3.101 falls in the critical region it is concluded that average level of the chemical is greater than 12.50. The null hypothesis is rejected. There is evidence that the chemical plant is in violation of the working code.

Step-by-step explanation:

Here

The population mean = u= 12.50

The sample mean = x`= 14.83

The sample size= n= 15

Sample standard deviation= 2.91

Let the null and alternate hypotheses be

H0  : u ≤ 12.50 against the claim that Ha: u > 12.50

Applying z- test

z= x`-u/s/√n

Z= 14.83-12.5/ 2.91/√15

Z= 2.33/ 0.751

z= 3.101

The significance level is chosen to be 0.05

The critical value of z at 0.05 > 1.645 for one tailed test

Since the calculated value of z= 3.101 falls in the critical region it is concluded that average level of the chemical is greater than 12.50. The null hypothesis is rejected. There is evidence that the chemical plant is in violation of the working code.

the answer is: the length is 5/6 yards.

give me a or a 5 star rating if this is correct!

step-by-step explanation:

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).