Answer:
4.72 hours/day
Step-by-step explanation:
Mean time spent watching TV (μ) = 2.8 hours a day
Standard deviation (σ) = 1.5 hours a day
The 90th percentile (upper 10%) of a normal distribution has an equivalent z-score of roughly z = 1.282. The minimum time spent watching TV, X, at the 90th percentile is:
On a typical day, you must watch at least 4.72 hours of TV to be in the upper 10%.
ANSWER TO QUESTION ONE IS 5
5x2+5=15
5-4=1
15x1=15
QUESTION 2
X = 20
IDK 3 :(
F = t ⇨ df = dt
dg = sec² 2t dt ⇨ g = (1/2) tan 2t
⇔
integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt
u = 2t ⇨ du = 2 dt
As integral of tan u = - ln (cos (u)), you get :
integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant
integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant
integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer
Answer:
The answer would be 155 tickets
