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jekas [21]
3 years ago
14

Divide the following polynomials, Write answer in descending powers of y. (2y to the third power + 3y squared - 4y - 5) (y + 1)

Mathematics
1 answer:
forsale [732]3 years ago
8 0
The first stwp for solving this expression is to multiply each term in the first parenthesis by each term in the second parenthesis (FOIL method).
2y³ × y + 2y³ + 3y² × y + 3y² - 4y × y - 4y - 5y - 5
Calculate the product of the first two numbers.
2y^{4} + 2y³ + 3y² × y + 3y² - 4y × y - 4y - 5y - 5
Calculate the product of the second multiplication expression. (2y² × y)
2y^{4} + 2y³ + 3y³ + 3y² - 4y × y - 4y - 5y - 5
Calculate the product of the final multiplication expression. (-4y × y)
2y^{4} + 2y³ + 3y³ + 3y² -4y² - 4y - 5y - 5
Collect the like terms that contain y³.
2y^{4} + 5y³ + 3y² -4y² - 4y - 5y - 5
Collect the like term that contain y².
2y^{4} + 5y³ + -y² - 4y - 5y - 5
Lastly,, collect the like terms that have y.
2y^{4} + 5y³ + -y² - 9y - 5
Since we cannot simplify this expression down any further,, the correct answer to your question is going to be 2y^{4} + 5y³ + -y² - 9y - 5.
Let me know if you have any further questions.
:)
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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.

8 0
3 years ago
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Andrews [41]

Answer:

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Step-by-step explanation:

3 0
3 years ago
K-10/2=7 two Step Equation
aivan3 [116]
I hope this helps you



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6 0
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Please answer this correctly without making mistakes I want ace expert and genius people to answer this correctly without making
soldi70 [24.7K]

Answer:

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Step-by-step explanation:

Ok so you are already the number that y = 10 and z = -2. You will plug these number in the equation in order to find the answer. :-

y = 10 and z = -2

yz + z

<u>10(-2)</u> - 2

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Hope this helps, thank you :) !!

6 0
3 years ago
A rectangular fence is 27.8 meters by 34.2 meters.what is its the perimeter answer
storchak [24]

The perimeter of a rectangle is calculated as

Perimeter= 2(length + breadth )

So, Perimeter = 2*length + 2*breadth

Now we are given the length as 34.2 meters and breadth as 27.8 meters

So the perimeter will be

2(34.2+27.8) meters

(2*34.2) + (2*27.8)

which equals 2(62) meters

Hence Perimeter = 124 meters

5 0
3 years ago
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