Two synchronized microwave sources. a and b. are emitting waves of wavelength 4.0 cm. source a is located on the x-axis atx-0, source b can be moved along the r-axis, and a microwave receiver is fixed on the r-axis at x 82 cm. for each of the following positions of source b determine whether the waves from the sources will reach the receiver in phase (constructive interference), out of phase (destructive interference), or with a phase difference of t 2 rad. 17% part (a) x=-8 cm. grade summary deductions potential ct/2 rad phase difference out of phase in phase 0% 100% submissions attempts remaining: 2 50% per attempt) detailed view submit hint hints: 1 % deduction per hint. hints remaining: 2 feedback: 0% deduction per feedback. 1700 part (b) x =-6 cm. 17% part (c) x =-5 cm. bb 17% part (d) x=-4 cm. 17% part ( cm. 17% part (f) x--2 cm. all content © 2019 expert ta

11.2

step-by-step explanation:

when doing tangent the tangent of the angle is equal to the opposite side from the angle over the adjacent side to the angle.

tan43 = x / 12

move 12 over by multiplying each side by 12

12tan43 = x

x = about 11.2

its totaly b

your welcome btw

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.