On each trial of an experiment, a subject is presented with a constant soft noise, which is interrupted at some unpredictable time by a noticeably louder sound.
The time it takes for the subject to react to this louder sound is recorded. The following list contains the reaction times (in milliseconds) for the 16 trials of this
experiment:
235, 222, 296, 152, 269,311, 164, 241, 186, 217, 261, 250, 202, 231, 273, 285
Find 25th and 90th percentiles for these reaction times.
(a)
The 25th percentile:
milliseconds
(b)
The 90th percentile:
milliseconds

168

step-by-step explanation:

i am pretty sure that the answer is 168

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).

answer is 8x+24

area of vanessa's patio = length × width

given, length = (3x² + 5x + 10)

width = (x² - 3x - 1)

area of vanessa's patio = (3x² + 5x + 10)(x² - 3x - 1)

= 3x²(x² - 3x - 1) + 5x(x² - 3x - 1) + 10(x² - 3x - 1)

= 3x⁴ - 9x³ - 3x² + 5x³ - 15x² - 5x + 10x² - 30x - 10

= 3x⁴ - 4x³ - 8x² - 35x - 10

hence, expression of vanessa's patio = (3x⁴ - 4x³ - 8x² - 35x - 10)

the answer for the question that you are asking is 4