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blondinia [14]
3 years ago
12

If I = prt; p = 350, r = 6%, I = 42, then t = 2 A. True B. False

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
4 0

t = 2 → A true

given I = PRT

to find T , divide both sides by PR

note that R = 6% = \frac{6}{100} = 0.06

T = \frac{I}{PR} = 42/( 350 × 0.06 ) = 2



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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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How much warmer is it<br>when the temperature is 17°F than<br>when it is -11°F?​
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Answer:

i thinks it -11°F but not sure

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The stem and leaf plot shows the number of items in each of the 12 collections in a museum.
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Step-by-step explanation:

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Given: (↔AB) ║ (↔CD)
nordsb [41]
1) m = y2 - y1/x2 - x1
y = mx + b
m = 7 - 0/3 - 8
y = -7/5x + 11.2
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8 0
2 years ago
Tomas and Tasha were hiking at a speed of 3.5 mph. On their way back, they were tired and their speed was 2 mph. How long was th
Anna35 [415]

 To get the length of the route use the expression for speed,

Distance = Speed × Time

Length of the route is 7 miles.

It's given in the question,

 Speed of Tomas and Tasha initially = 3.5 mph

Therefore, distance travelled by both with the speed of 2 mph one way will be,

 Distance = 3.5t [Here, t = time ]

Speed of both while coming back = 2 mph

And the time taken = t + (1 hour 30 minutes)

                                 = (t + 1.5) hours

Therefore, distance travelled while coming back = Speed × Time

                                                                                  = 2(t + 1.5)

Since, length of the route is same both ways,

3.5t = 2(t + 1.5)

3.5t = 2t + 3

3.5t - 2t = 3

1.5t = 3

t = 2 hours

Now substitute the value of t in the expression to get the distance of the route,

Distance = 3.5t

               = 3.5 × 2

               = 7 miles

 Therefore, route was 7 miles long.

Learn more,

brainly.com/question/1955818

7 0
2 years ago
Read 2 more answers
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