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lukranit [14]
2 years ago
6

How can you write the expression 7(b – 2) in words?

Mathematics
2 answers:
mash [69]2 years ago
6 0

Answer:

Option E

Step-by-step explanation:

Let's check

  • Difference of b-2
  • It has multiplied with 7

So the expression becomes 7(b-2)

m_a_m_a [10]2 years ago
3 0

Solution:

Let's verify each option to see which is correct.

<u>Option A</u>

Two subtracted from the quotient of seven divided by b.

  • => 7/b - 2 = 7(b - 2)                                                                [False]

<u>Option B</u>

Seven added to difference of b minus two.

  • => 7 + (b - 2) = 7(b - 2)                                                           [False]

<u>Option C</u>

The quotient of seven divided by b minus two.

  • => 7/b - 2 = 7(b - 2)                                                               [False]

<u>Option D</u>

Two subtracted from seven times b

  • => 7b - 2 = 7(b - 2)                                                                [False]

<u>Option E</u>

The product of seven and the difference of b minus two

  • => 7 x (b - 2) = 7(b - 2) = 7(b - 2)                                           [True]

Option E is correct.

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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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Answer:

(4, 1)

Step-by-step explanation:

For any rectangle or square located on the axes, match the pairs and find what's missing. 3 -- 3, 4 -- 4. unmatched are 4 and 1. 4 is x because it's missing from x and same for 1 and y.

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Answer:

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Step-by-step explanation:

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4 0
3 years ago
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Andrew [12]

Answer:

13 \frac{1}{3} c - 1 \frac{1}{2}

Step-by-step explanation:

We want to combine alike terms so

-2/3c+14c & -9/5+3/10

14c-2/3

\frac{14}{1} c -  \frac{2}{3} c

Multiply 14/1 by 3/3 to get the denominators the same

\frac{42}{3} c -  \frac{2}{3} c

Subtract

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simplify

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now

\frac{3}{10}  -  \frac{9}{5}

Multiply 9/5 by 2/2

\frac{3}{10}  -  \frac{ 18}{10}

Subtract

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Simplify

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put both together

13 \frac{1}{3}c  - 1 \frac{1}{2}

Hope this helps! If you have any questions on how I got my answer feel free to ask. Stay safe!

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