Answer:
How many standard deviations above the mean is 14,500 hours? 1.25 1.5 2.5 Using the standard normal table, the probability that Seth's light bulb will last no more than 14,500 (P(z ≤ 1.25)) hours is about ✔ 89% .
Answer:
a) The data distribution consists of ( 7 )1's (denoting a foreign student) and ( 43 )0's (denoting a student from the U.S.).
b) The population distribution consists of the x-values of the population of 12,152 full-time undergraduate students at theuniversity, ( 6 )% of which are 1's (denoting a foreign student) and ( 94 )% of which are 0's (denoting a student from the U.S.).
c) The mean is ( 0.06 )
The standard deviation is ( 0.0336 )
The sampling distribution represents the probability distribution of the ( sample ) proportion of foreign students in a random sample of ( 50 ) students. In this case, the sampling distribution is approximately normal with a mean of ( 0.06 ) and a standard deviation of ( 0.0336 )
Step-by-step explanation:
58% is about equal to 60%, 10% of 121 is 12.1.
12.1 * 6 = 72.6
To take this further, 58% is 2% less than 60%, or 2 times 1%. 1% of 121 is 1.21.
72.6 - (1.21 * 2) = 72.6 - 2.42 = 70.18
To evaluate 17 int (sin^2 (x) cos^3(x))
From Trig identity. Cos^2(x) + sin^2(x) =1. Cos^2(x) = 1 - sin^2 (x)
Cos^3(x) = cosx * (1 - sin^2 (x)) = cosx - cosxsin^2x
So we have 17 int (sin^2x(cosx - cosxsin^2x))
int (sin^2x(cosx)dx - int (sin^4xcosx)dx. ----------(1)
Let u = sinx then du = cosxdx
Substituting into (1) we have
int (u^2du) - int (u^4du)
u^3/3 - u^5/5
Substitute value for u we have
(sinx)^3/3 - (sinx)^5/5
Hence we have 17 [ sin^3x/3 - sin^5x/5]
Let x be the width of the playground, then 3x is the length of the
<span>playground
х * 3х = 75
3x</span>² = 75
x² = 25
x = 5 m (width)
5*3=15 m (length)
Perimeter = 2(5+15) = 2*20 = 40 meters.
