Answer:
Z scores of -2 or lower are considered unusually low. Since the z-score of a 49-cm head circunference is -2, it is an unusual measure.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Z scores of -2 or lower are considered unusually low, and zscores of 2 or higher are considered unusually high.
In this problem, we have that:
49cm head circunference unusual?
Z scores of -2 or lower are considered unusually low. Since the z-score of a 49-cm head circunference is -2, it is an unusual measure.
256×7=1792 in expanded form it is 1000+700+90+2
Question:
A certain vibrating system satisfies the equation u''+γu'+u=0. Find the value of the damping coefficientγfor which the quasi period of the damped motion is 50% greater than the period of the corresponding undamped motion.
Answer: y = √(20/9) = √20/3 = 1.49071
Step-by-step explanation:
u''+γu'+u=0
m =1, k =1, w• = √ (k/m) = 1
The period of undamped motion T, is given by T = 2π/w•, T = 2π/1 = 2π
The quasi period Tq = 2π/quasi frequency
Quasi frequency = ((4km - y^2)^1/2)/2m
Therefore the quasi period Tq = 4πm/((4km - y^2)^1/2)
From the question the quasi period is 50% greater than the period of undamped motion
Therefore Tq = T + (1/2)T = (3/2)T
Thus,
4πm/((4km - y^2)^1/2) = (3/2)(2π)
Where, k =1, m=1,
4π/((4 - y^2)^1/2) = 3π,
(4 - y^2)^1/2 = 4π/3π,
(4 - y^2) = (4/3)^2,
4 - y^2 = 16/9,
y^2 =4 - 16/9,
y^2 = 20/9,
y = √(20/9)
The next number would be 20 because every other number is just multiplying by 2.
Ex.
3, 6. 3•2=6
4, 8. 4•2=8