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olasank [31]
3 years ago
12

Chapter 7 of the Jiuzhang suanshu presents a problem of two linear equations involving acres of land and their respective prices

. One of the two equations can be translated to: 300x + 300 x plus StartFraction 500 over 7 EndFraction left-parenthesis 87.5 right-parenthesis equals 10,000.y = 10000 If y = 87.5, what is the value for x? 300x + 300 x plus StartFraction 500 over 7 EndFraction y equals 10,000.y = 10,000 300x + 300 x plus StartFraction 500 Over 7 EndFraction y equals 10,000.(87.5) = 10,000 300x + 6,250 = 10,000 x =
Mathematics
2 answers:
Nutka1998 [239]3 years ago
7 0

Answer:

12.5

Step-by-step explanation:

i just did it

iVinArrow [24]3 years ago
6 0

Answer: 12.5

Step-by-step explanation:

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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 s
sp2606 [1]

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)

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0
2 years ago
6 × (9 + 9)
Aleksandr [31]

Answer:

C

Step-by-step explanation:

Becuz 9+9=17, which is the sum of 9 and 9.

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C. Associative property of communication.
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What can you conclude about the setting, based on this passage? Buck is in the north, where sled dogs are in demand. Buck is exp
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Answer:

We can conclude that the setting is stationed somewhere in the north where it's very cold, so cold that they can use sled dogs. We can also assume it's winter time because it normally snows during the winter.

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