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Archy [21]
1 year ago
13

Kaitlin, Tony, and Austin sent a total of 103 text messages over their cell phones during the weekend. Austin sent 4 times as ma

ny messages as Tony. Tony sent 5 more messages than Kaitlin. How many messages did they each send?
Mathematics
1 answer:
emmainna [20.7K]1 year ago
3 0

The number of messages Kaitlyn, Tony and Austin sent are 13, 18 and 72 messages respectively.

<h3>How to write and solve equation?</h3>

let

  • Number of messages Kaitlyn sent = x
  • Number of messages Tony sent = x + 5
  • Number of messages Austin sent = 4(x + 5)
  • Total messages = 103

103 = x + (x + 5) + 4(x + 5)

103 = x + x + 5 + 4x + 20

103 = 6x + 25

103 - 25 = 6x

78 = 6x

x = 78/6

x = 13

Number of messages each sent:

Number of messages Kaitlyn sent = x

= 13 messages

Number of messages Tony sent = x + 5

= 13 + 5

= 18 messages

Number of messages Austin sent = 4(x + 5)

= 4(13 + 5)

= 4(18)

= 72 messages

Learn more about equation:

brainly.com/question/2972832

#SPJ1

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We are given that a group of researchers at the BYU Off-Campus Housing department want to estimate the mean monthly rent that unmarried BYU students paid during winter 2018.

During March 2018, they randomly sampled 314 BYU students and found that on average, students paid $346 for rent with a standard deviation of $86.

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where, \bar X = average rent paid by 314 BYU students = $346

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So, 96% confidence interval for the population mean monthly rent, \mu is ;

P(-2.082 < t_3_1_3 < 2.082) = 0.96

P(-2.082 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 2.082) = 0.96

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P( \bar X -2.082 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X +2.082 \times {\frac{s}{\sqrt{n} } } ) = 0.96

96% confidence interval for \mu = [ \bar X -2.082 \times {\frac{s}{\sqrt{n} } } , \bar X +2.082 \times {\frac{s}{\sqrt{n} } } ]

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Therefore, 96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

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