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evablogger [386]
2 years ago
5

The speed of a sparrow is x km/h in still air. When the

Mathematics
1 answer:
Semmy [17]2 years ago
4 0

Answer:

x = 5m/s

Step-by-step explanation:

Distance flying out = 12 km  (headwind)

Distance flying back = 12 km (tailwind)

total distance = 12 + 12 =24 km

wind speed = 1km/h

speed going out (with headwind) = (x - 1) km/h

speed coming back (with tailwind) = (x + 1) km/h

Time taken to go out = distance going out / speed going out

= 12 / (x-1)

Time taken to come back = distance coming back / speed coming back

= 12 / (x+1)

total time = time taken to go out + time taken to come back

5 =[ 12/(x-1) ] + [ 12/(x-1)]

expanding this, we will get

5x² - 24x - 5 = 0

solving quadratic equation, we will get

x = -1/5 (impossible because speed cannot be negative)

or

x = 5 (answer)

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2 years ago
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2 years ago
A process engineer is implementing a quality assurance system on a breakfast cereal production line. A new sensor is installed o
nataly862011 [7]

Answer:

0.2%

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The given parameters are;

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\begin{array}{cccc}&P&N&T\\P&TP&0.9&90\\N&FP&9.8&10\end{array}

We have, FN = 0.9, TN = 9.8, TP = 90 - 0.9 = 89.1, FP = 10 - 9.8 = 0.2

FNR = FN/(FN + TP) = 0.9/(0.9 + 89.1) = 0.01

FPR = 0.2/(0.2 + 9.8) = 0.02

TPR = 89.1/(89.1 + 0.9) = 0.99

TNR = 9.8/(9.8 + 0.2) = 0.98

P(C/A) = TPR × P(C)/((TPR × P(C) + FPR×(1 - P(C)))

Where;

P(C/A) = The probability that a correct weight package is accepted

∴ P(C/A) = 0.99*0.9/(0.99*0.9 + 0.02*0.1) ≈ 0.99776

The probability that a correctly weighed package is rejected, P(C/R), is given as follows;

P(C/R) = 1 - P(C/A)

∴ P(C/R) ≈ 1 - 0.99776 = 0.00224 ≈ 0.2%

The probability that a correctly weighed package is rejected, P(C/R) ≈ 0.2%

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Answer:

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