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kifflom [539]
3 years ago
8

The total ages of Sarah, Jake and Matthew add up to 36. Jake is two years younger than Sarah and Matthew is 16 years younger tha

n Sarah. If you let the variable
represent Sarah's age then which of the following would best represent the ages of:
Jake
and Matthew
O A. Jake: x + 4)
and Matthew: x-8)
OB Jake
x + 2
and Matthew: x - 16)
O Jake -2)
and Matthew: x - 16)
OD Jake: x+2
and Matthew: X16)
Next >
< Previous
go
D
DD
so
Mathematics
2 answers:
Hatshy [7]3 years ago
6 0

Answer:

Step-by-step explanation:

S+J+M=36

J=S-2 or J=x-2

M=S-16 or M=x-16

Natalija [7]3 years ago
6 0

Answer:

Jake: x-2 and Matthew: x-16  (I think it is C on your sheet.)

Step-by-step explanation:

x = Sarah

x-2 = Jake

x-16 = Matthew

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You spin the spinner and flip a coin. Find the probability of the compound event.
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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
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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)

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
3 years ago
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