The American Psychological Associated conducted a survey in 2003 of psychologists to estimate the average incomes for psychologists based on their academic degree and years of experience. 36 psychologists responded who had a just received their masters degree in 2003. The average was $48,800 with a standard deviation of $16,800.

This question is incomplete, the complete question;

The American Psychological Associated conducted a survey in 2003 of psychologists to estimate the average incomes for psychologists based on their academic degree and years of experience. 36 psychologists responded who had a just received their masters degree in 2003. The average was $48,800 with a standard deviation of $16,800.

The 95% Confidence Interval is ($, $ )

at 95%, confidence interval is ( $43312, $54288 )

Step-by-step explanation:

Given the data in the question;

sample size n = 36

mean x' = 48,800

standard deviation σ = 16,800

at 95% confidence level;

∝ = 1 - 95% = 1 - 0.95 = 0.05

∝/2 = 0.05 / 2 = 0.025

z critical value, = 1.96

now, confidence interval for population mean is given by the formula

x' ±  ( σ/√n )

so we substitute

⇒ 48800 ±  1.96( 16800/√36 )

⇒ 48800 ±  1.96( 2800 )

⇒ 48800 ±  5488

⇒ 48800 -  5488, 48800 +  5488

⇒ ( 43312, 54288 )

Therefore, at 95%, confidence interval is ( $43312, $54288 )

i got 86 for this

hope this !

if you need an explanation just ask

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)

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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

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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).