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elixir [45]
3 years ago
11

A shark weighs 405 kg and 68 grams a second shark weighs 324 kg and 75 g how much more does the first shark weigh in grams than

the second shark
Mathematics
1 answer:
mariarad [96]3 years ago
8 0

For this case we have to by definition:

1 kg equals 1000 grams

Shark 1:

405 kg and 68 grams

405 kg * \frac {1000 g} {1kg} = 405,000 grams

Thus, shark 1 weighs 405,068 grams.

Shark 2:

324 kg and 75 grams

324 kg * \frac {1000 g} {1 kg} = 324,000 grams

Thus, shark 2 weighs 324,075 grams.

Subtracting we have:

405.068 grams-324.075 grams = 80.993 grams

Thus, shark 1 weighs 80,993 grams more than the second.

Answer:

Shark 1 weighs 80,993 grams more than the second.

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Answer:

(1) Variance = 4.5 and Standard deviation = 2.121.

(2) Variance = 4.5 and Standard deviation = 2.121.

(3) The effect on the measure of dispersion if each value is changed uniformly ​is that it remains unchanged.

Step-by-step explanation:

We are given with the following set of data below;

              X                     X-\bar X                   (X-\bar X)^{2}

              5                   5 - 8 = -3                     9

              5                   5 - 8 = -3                     9

              8                   8 - 8 = 0                      0

             10                  10 - 8 = 2                     4

             10                  10 - 8 = 2                     4

             10                  10 - 8 = 2                     4

              9                   9 - 8 = 1                       1

              9                   9 - 8 = 1                       1

         <u>     6     </u>              6 - 8 = -2             <u>        4       </u>          

Total <u>    72     </u>                                         <u>       36       </u>

<u>Firstly, the mean of the above data is given by;</u>

              Mean, \bar X  =  \frac{\sum X}{n}

                              =  \frac{72}{9}  = 8

(1)Now, the variance of the given data is;

            Variance  =  \frac{\sum (X-\bar X)^{2} }{n-1}

                                  =  \frac{36}{9-1}  = 4.5

So, the standard deviation, (S.D.)  =  \sqrt{\text{Variance}}

                                                        =  \sqrt{4.5} = 2.12

(2) Now, each value is added by 2; so the new data set is given by;

              X                     X-\bar X                   (X-\bar X)^{2}

              7                   7 - 10 = -3                     9

              7                   7 - 10 = -3                     9

             10                  10 - 10 = 0                     0

             12                  12 - 10 = 2                     4

             12                  12 - 10 = 2                     4

             12                  12 - 10 = 2                     4

              11                  11 - 10 = 1                       1

              11                  11 - 10 = 1                       1

         <u>     8     </u>              8 - 10 = -2             <u>        4       </u>          

Total <u>    90     </u>                                         <u>       36       </u>

<u>Firstly, the mean of the above data is given by;</u>

              Mean, \bar X  =  \frac{\sum X}{n}

                              =  \frac{90}{9}  = 10

(1)Now, the variance of the given data is;

            Variance  =  \frac{\sum (X-\bar X)^{2} }{n-1}

                                  =  \frac{36}{9-1}  = 4.5

So, the new standard deviation, (S.D.)  =  \sqrt{\text{Variance}}

                                                                =  \sqrt{4.5} = 2.12

(3) The effect on the measure of dispersion if each value is changed uniformly ​is that it remains unchanged as we see in the case of variance or standard deviation.

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