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krek1111 [17]
3 years ago
6

A police car's siren operates at a frequency of 2,000 Hz. What frequency would a stationary listener observe if the police car i

s driving towards him at 65 mph? Note vsound = fλ, where vsound (the speed of sound in air) is 345 m/s. Also, 1 mile = 1.609 km.
Mathematics
1 answer:
SpyIntel [72]3 years ago
3 0

we can use formula

f_L=f_s(\frac{v_a-v_p}{v_a+v_p} )

where

fL is listener frequency

fs is siren frequency

va is the speed of sound in air

vs is police car speed

so, we have

f_s=2000Hz

v_a=345m/s

v_p=65mph

now, we can change it into m/s

v_p=65*\frac{1609}{3600}m/s

v_p=29.05139m/s

now, we can plug it

f_L=2000(\frac{345-29.05139}{345+29.05139} )

f_L=1689.33263Hz..............Answer

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Data : 32.1 , 30.6 , 31.4 , 30.4 , 31.0 , 31.9

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= (31.23 - 31) /(0.65/sqrt(6))

= 0.867

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z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.65/sqrt(6))

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i) SD = 0.8

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z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.8/sqrt(6))

= 0.704

P-value = 0.4814 Answer

ii) SD = 0.8

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z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.8/sqrt(6))

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P-value = 0.0184 Answer

The probabilities obtained in part c are comparatively higher than that of part b.

d)

i) For alpha =0.01

z = (Xbar - mu) / (SD/sqrt(n))

=> -2.32 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-2.32

=> sqrt(n) = (0.65*(-2.32)) / (31.23 - 31)

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ii) For beta =0.10

z = (Xbar - mu) / (SD/sqrt(n))

=> -1.28 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-1.28

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=> n = 13 Answer

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3 years ago
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