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taurus [48]
3 years ago
11

Sharon is 54 inches tall. tree in her backyard is five times as tall as she is. the floor of her treehouse is at a height that i

s twice as tall as she is. What is the difference in inches between the top of the tree and the floor of the treehouse?
Mathematics
2 answers:
mafiozo [28]3 years ago
3 0
First to solve this problem you are going to figure out how tall the tree is. So if Sharon is 54 inches and the tree 5 times the height she is you would do 54*5= 270. Next you go to find out how tall the floor of her tree house is. It says the floor is twice the height of her so you would do 54*2=108. Then the problem asks the difference between the top of the tree and the floor of the tree house. So 270-108=162. I hope this helped. :) Brainliest answer?
chloe sexy needs bf 2 years ago
0 0

270 (54x5)

chloe sexy needs bf
2 years ago
I will be ur bf
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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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stealth61 [152]

Answer:

(x-10)° + (x + 14)° = 180°

=> x° - 10° + x° + 14 ° = 180°

=> 2x° + 4° = 180°

=> 2x° = 176°

=> x° = 176°/ 2

=> x° = 88°

Therefore the magnitude of x° = 88° (ans)

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Answer:

There are 25 students in the class.

Step-by-step explanation:

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Let g be the function given by g(x)=-3x^2-2x+3 evaluate g(-2)
sashaice [31]

g(x) = -3x² - 2x + 3

g(-2) = -3(-2)² - 2(-2) + 3

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Answer:

A

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