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disa [49]
3 years ago
14

paloma eats 2/8 of an apple in the morning and 3/8 more in the afternoon. how much of an apple did she eat in total

Mathematics
1 answer:
marishachu [46]3 years ago
6 0

Step-by-step explanation:

in this problem, we are solving for the amount of apple that Paloma eats

1. she ate 2/8 of a complete apple -hence the will be a remainder, we will find  

  that remainder

2. she also ate 3/8 of the remainder for lunch

3. we then proceed to calculate all apple she ate in total

1. \frac{2}{8} *1= \frac{1}{4} \\\\1-\frac{1}{4}= \frac{4-1}{4}  = \frac{3}{4}

2 she ate 3/8 of the remaining 3/4 apple

1. \frac{3}{8} *\frac{3}{4} = \frac{9}{32}

3. in  total she ate 1/4+ 9/32

\frac{1}{4} +\frac{9}{32} = \frac{8+9}{32} = \frac{17}{32}

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Ne4ueva [31]

Answer:

The test statistics is  t  =  -1.727

Step-by-step explanation:

From the question we are told that

The data given is  

   330 620 1870 2410 4620 6396 7822 81028309 12882 14419 16092 18384 20916 23812 25814

 The population mean is  \mu  =  14400

    The  sample  size is  n =  16

  The  null hypothesis is  \mu \le  14400

    The  alternative hypothesis is  H_a  :  \mu > 14400

The sample mean is mathematically evaluated as

  \= x  =  \frac{\sum x_i}{n}

So

   \= x  =  \frac{330+ 620+ 1870 +2410+ 4620+ 6396+ 7822+ 8102+8309+ 12882+ 14419+ 16092+ 18384 +20916+ 23812+ 25814 }{16}

=>  \= x = 10799.9

The  standard deviation is mathematically represented as

      \sigma =\sqrt{\frac{ \sum (x_i - \=x)^2}{n}}

So

\sigma =\sqrt{\frac{(330- 10799.9)^2 + (620- 10799.9)^2+ (1870- 10799.9)^2 +(2410- 10799.9)^2 + (4620- 10799.9)^2 +(6396- 10799.9)^2 +(7822- 10799.9)^2 }{16}}  \ ..

   ..\sqrt{ \frac{(8102 - 10799.9)^2 +(8309 - 10799.9)^2 + (12882 - 10799.9)^2 + (14419 - 10799.9)^2 + (16092 - 10799.9)^2 + (18384 - 10799.9)^2 +(20916 - 10799.9)^2  }{16}} \ ...

  \ ... \sqrt{\frac{(23812 - 10799.9)^2 +(25814 - 10799.9)^2 }{16}}

=>  \sigma  =  8340

  Generally the test statistic is mathematically represented as

  t =  \frac{10799.9- 14400}{ \frac{8340}{\sqrt{16} } }

t  =  -1.727

From the z-table  the p-value is  

     p-value  = P(Z > t) =  P(Z >  -1.727) =  0.95792

 From the values obtained we see that

        p-value  >  \alpha  so we fail to reject the null hypothesis

Which implies that the claim of the NarStor is wrong

5 0
3 years ago
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