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4 years ago

3.4k+(0.63−0.81k)÷0.9=5.7

Mathematics

2 answers:

Monica [59]4 years ago

5 0

The answer is k = 2

iren2701 [21]4 years ago

5 0

Answer:

k = 1.93 which can be rounded up to 2

Step-by-step explanation:

3.4k + (0.63 − 0.81k) ÷ 0.9 = 5.7 You don't need parentheses because you can't distribute

3.4k + 0.63 − 0.81k ÷ 0.9 = 5.7 Combine like terms

2.59k + 0.7 = 5.7

- 0.7 - 0.7 Subtract 0.7 from both sides

2.59k = 5 Divide both sides by 2.59

k = 1.93 which can be rounded up to 2

If this answer is correct, please make me Brainliest!

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Step-by-step explanation:

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2 years ago

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GenaCL600 [577]

Answer:

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Step-by-step explanation:

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Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

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c) Less than 7.5 minutes:

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NORMSDIST (-3.92)

= 0.0000

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3 years ago