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andrew11 [14]
3 years ago
14

The manager of a baseball team has 15 players to choose from for his 9 person batting order

Mathematics
2 answers:
Masteriza [31]3 years ago
6 0
Given the question:
The manager of a baseball team has 15 players to choose from for his 9-person batting order. How many different ways can he arrange the players in the lineup?
5,005    362,880    3,603,600    1,816,214,400

The number of was in which a smaller group is choosing from a bigger group is given by the bigger number combination the smaller number.
^nC_r= \frac{n!}{(n-r)!r!}
Here, we want to choose a 9-person batting order from a team of 15 players. This is given by:
^{15}C_9= \frac{15!}{(15-9)!9!}=\frac{15\times14\times13\times12\times11\times10\times9!}{6!\times9!}= \frac{3,603,600}{720} =5,005

omeli [17]3 years ago
3 0

Answer:

The answer is 1,816,214,400 not 5005. Just took the quiz


Step-by-step explanation:


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Can someone help me solve for x in this equation?? 25+3.50x < 17+7.50x
Lelu [443]
So we want to solve like if it was just a basic equation. We start with 25+3.50x<17+7.50x and get rid of our smallest x. That means we get rid of the 3.50x by subtracting from both sides and our new equation is 25<17+4x. Let’s get our x by itself shall we, so let’s subtract 17 from both sides and we get 8<4x. Finally all we do is divide by 4 and just like that 2
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3 years ago
The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)
Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1 .

Step-by-step explanation:

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

Let A = Math Scores ,B = Writing Scores  and D = difference between both

So, \mu_A = Population mean for the math scores

       \mu_B = Population mean for the writing scores

 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

     8                       499                           459                                   -40

     9                       610                            615                                       5

     10                      572                           541                                      -31

     11                       390                           335                                     -55

     12                      593                           613                                       20  

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

 Bbar = \frac{\sum B_i}{n} = \frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}  = 489

 Abar =  \frac{\sum A_i}{n} = \frac{540+432+528+574+448+502+480+499+610+572+390+593}{12} = 514

 ∑D_i^{2} = 22600     and  s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}} = \sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} } = 37.05

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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elixir [45]
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daser333 [38]

Answer:

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Step-by-step explanation:

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