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photoshop1234 [79]
3 years ago
14

Bharat and ingrid leave their building at the same time on their bikes and travel in opposite directions. If bharat's speed is 1

2 kilometers per hour and the Ingrid's speed is 14 kilometers per hour, how long will it take until they are 78 kilometers apart ?
Mathematics
1 answer:
yanalaym [24]3 years ago
6 0

Answer:

3 hours

Step-by-step explanation:

speed of Bharat = 12kilometers / hour

speed of Ingrid=14 kilometers / hour

For this problem we'll be using formula relating time, distance and speed.i.e

Distance = speed x time

Suppose Bharat is riding at a distance of 'b' kilometers, therefore time taken by him will be:

Time = distance/ speed

Time= b/12 hours.

Also, Ingrid can ride the distance at this time with speed of 14km/hr

The distance would be,

Distance = speed x time

Distance = 14 x (b/12) => 14b/12

Distance= 7b/6

Let 'd_{b}' be the distance that Bharat have covered after time 't'

therefore, d_{b} = 12 x t

Let 'd_{i} ' be the distance that Ingrid have covered after time 't'

therefore, d_{i} = 14 x t

In order to find the time when they are 78 kilometers apart, we will add d_{i}  and d_{b}, because they are travelling in opposite direction creating distance between them.

So,

d_{i} + d_{b} = 78

( 14 x t) + (12 x t) =78

14t + 12t =78

26t= 78

t= 78/26

t= 3hours.

thus, it will take 3 hours  until they are 78 kilometers apart

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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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