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lozanna [386]
3 years ago
9

In January Jennifer was allowed to use 150 text messages. If she is allowed to use the same amount of text messages every month.

How many text messages would she have used in all by the month of August?
Mathematics
1 answer:
Firlakuza [10]3 years ago
6 0

She would have used 1200 messages in all by the month of August.

Step-by-step explanation:

Messages allowed per month = 150

There will be 8 months starting from January to August, therefore,

Time period = 8 months

Messages in 8 months = Messages per month * Time period

Messages\ in\ 8\ months=150*8\\Messages\ in\ 8\ months=1200

She would have used 1200 messages in all by the month of August.

Keywords: Multiplication

Learn more about multiplication at:

  • brainly.com/question/629998
  • brainly.com/question/6208262

#LearnwithBrainly

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46,125

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onsider the following hypothesis test: H 0: 50 H a: > 50 A sample of 50 is used and the population standard deviation is 6. U
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Answer:

a) z(e)  >  z(c)   2.94 > 1.64  we are in the rejection zone for H₀  we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) z(e) < z(c)  1.18 < 1.64  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

c) 2.12  > 1.64 and we can conclude the same as in case a

Step-by-step explanation:

The problem is concerning test hypothesis on one tail (the right one)

The critical point  z(c) ;  α = 0.05  fom z table w get   z(c) = 1.64 we need to compare values (between z(c)  and z(e) )

The test hypothesis is:  

a) H₀      ⇒      μ₀  = 50     a)  Hₐ    μ > 50   ;    for value 52.5

                                          b) Hₐ    μ > 50   ;     for value 51

                                          c) Hₐ    μ > 50   ;      for value 51.8

With value 52.5

The test statistic    z(e)  ??

a)  z(e) =  ( μ  -  μ₀ ) /( σ/√50)      z(e) = (2.5*√50 )/6   z(e) = 2.94

2.94 > 1.64  we are in the rejected zone for H₀  we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) With value 51

z(e) =  ( μ  -  μ₀ ) /( σ/√50)    ⇒  z(e) =  √50/6    ⇒  z(e) = 1.18

z(e) < z(c)  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

c) the value 51.8

z(e)  =  ( μ  -  μ₀ ) /( σ/√50)    ⇒ z(e)  = (1.8*√50)/ 6   ⇒ z(e) = 2.12

2.12  > 1.64 and we can conclude the same as in case a

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If MNO ~XYZ,then what corresponding parts are congruent
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Answer:

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Step-by-step explanation:

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