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icang [17]
2 years ago
13

An airplane is flying at a speed of 580 km/h on a bearing of North 60° east. The wind blows from a bearing of North 45° west at

a speed of 60 km/h. What is the planes actual ground speed kilometers per hour? What is the planes true course? Write the true direction as an angle theta from the positive X axis.
Mathematics
1 answer:
kiruha [24]2 years ago
5 0

Answer:

t

Step-by-step explanation:

yhhhhh6bbbb h 66yltktktktoitt

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Approximately 1.143 L of peroxide should be added to the original solution. 
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Solve and type in a different form by using the given theorems of logarithms. <br> log N^3
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Log N^3  = 3 log N

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An athletic field is a 48 yd​-by-96 yd ​rectangle, with a semicircle at each of the short sides. A running track 20 yd wide surr
MA_775_DIABLO [31]

The distance would be equal to 2 times the longest side of the rectangle plus twice the shortest side multiplied by pi / 2 for the semicircle, that is:

longest side 96 and shortest 48

D = 2 * (96) + 2 * (1/2) * pi * 48

D = 192 + pi * 48

This shorter side, which starts at 48, will expand each time by two more in proportion to 20 of the running track between 8 than the number of divisions, that is, 2 * (20/8) = 5

In other words, there are 8 distances, like this:

D1 = 192 + 3.14 * 48 = 342.72 yd

D2 = 192 + 3.14 * (48 + 5) = 358.42 yd

D3 = 192 + 3.14 * (48 + 10) = 374.12 yd

D4 = 192 + 3.14 * (48 + 15) = 389.82 yd

D5 = 192 + 3.14 * (48 + 20) = 405.52 yd

D6 = 192 + 3.14 * (48 + 25) = 421.22 yd

D7 = 192 + 3.14 * (48 + 30) = 436.92 yd

D8 = 192 + 3.14 * (48 + 35) = 452.62 yd

6 0
3 years ago
The diameters of ball bearings are distributed normally. The mean diameter is 87 millimeters and the standard deviation is 6 mil
devlian [24]

Answer:

69.14% probability that the diameter of a selected bearing is greater than 84 millimeters

Step-by-step explanation:

According to the Question,

Given That, The diameters of ball bearings are distributed normally. The mean diameter is 87 millimeters and the standard deviation is 6 millimeters. Find the probability that the diameter of a selected bearing is greater than 84 millimeters.

  • In a set with mean and standard deviation, the Z score of a measure X is given by Z = (X-μ)/σ

we have μ=87 , σ=6 & X=84

  • Find the probability that the diameter of a selected bearing is greater than 84 millimeters

This is 1 subtracted by the p-value of Z when X = 84.

So, Z = (84-87)/6

Z = -3/6

Z = -0.5 has a p-value of 0.30854.

⇒1 - 0.30854 = 0.69146

  • 0.69146 = 69.14% probability that the diameter of a selected bearing is greater than 84 millimeters.

Note- (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 p-value, we get the probability that the value of the measure is greater than X)

7 0
3 years ago
PLEASE NEED HELP
expeople1 [14]

Answer:

D) 96%

Step-by-step explanation:

If you take 96% of 28.6 you get 27.6 so it fits

4 0
2 years ago
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