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Vitek1552 [10]
3 years ago
11

At Katya's fruit stand, a basket of strawberries costs $4 and a basket of raspberries costs $9. In one morning, Katya sells 96 b

askets for $644. How many baskets of strawberries were sold
Mathematics
1 answer:
Cloud [144]3 years ago
6 0

44 baskets of strawberries are sold

<em><u>Solution:</u></em>

Let "a" be the number of baskets of strawberries sold

Let "b" be the number of raspberries sold

Cost of 1 basket of strawberries = $ 4

Cost of 1 basket of raspberries = $ 9

<em><u>96 baskets were sold. Therefore, we can say,</u></em>

number of baskets of strawberries + number of raspberries sold = 96

a + b = 96

b = 96 - a ------------- eqn 1

<em><u>In one morning, Katya sells 96 baskets for $644</u></em>

Thus we frame a equation as:

a \times 4 + b \times 9 = 644

4a + 9b = 644 ----------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

<em><u>Substitute eqn 1 in eqn 2</u></em>

4a + 9(96 - a) = 644

4a + 864 - 9a = 644

5a = 864 - 644

5a = 220

Divide both sides by 5

a = 44

Thus 44 baskets of strawberries are sold

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We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

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So, \mu_A = Population mean for the math scores

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 Let \mu_D = Difference between the population mean for the math scores and the population mean for the writing scores.

            <em>  Null Hypothesis, </em>H_0<em> : </em>\mu_A = \mu_B<em>     or   </em>\mu_D<em> = 0 </em>

<em>      Alternate Hypothesis, </em>H_1<em> : </em>\mu_A \neq  \mu_B<em>      or   </em>\mu_D \neq<em> 0</em>

Hence, Test Statistics used here will be;

            \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1    where, Dbar = Bbar - Abar

                                                               s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}

                                                               n = 12

Student        Math scores (A)          Writing scores (B)         D = B - A

     1                      540                            474                                   -66

     2                      432                           380                                    -52  

     3                      528                           463                                    -65

     4                       574                          612                                      38

     5                       448                          420                                    -28

     6                       502                          526                                    24

     7                       480                           430                                     -50

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     10                      572                           541                                      -31

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Now Dbar = Bbar - Abar = 489 - 514 = -25

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So, Test statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1

                            = \frac{-25 - 0}{\frac{37.05}{\sqrt{12} } } follows t_1_1   = -2.34

<em>Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .</em>

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

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