A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 60 cookies. The mean is 22.36 and the standard deviation is2.97 . Construct a ​80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

2.665 < σ < 3.379

Step-by-step explanation:

Given :

s = 2.97

Sample size, n = 60

α = 80%

χ² Critical value (two - tailed), df = (60-1) = 59

χ² = 45.577 ; χ² = 73.279

The 80% confidence interval for the standard deviation :

s * √(n - 1) / χ² critical

2.97 * √(60 - 1) / 73.279 < σ < 2.97 * √(60 - 1) / 45.577

2.665 < σ < 3.379

answer:

c. the third graph

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752