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grandymaker [24]
2 years ago
6

Your break even point is 3000 units you make $5 each unit that is your price minus variable coals is $5 what your fixed cost ? U

se the formula break even points = fixed cost / price
Mathematics
1 answer:
Goshia [24]2 years ago
5 0

Answer: ummm i think do times or -

Step-by-step explanation:

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What’s the answer? Help
iren2701 [21]

Answer:

$14

Step-by-step explanation:

x = price of scarf 1

y = price of scarf 2

together they cost $25

x + y = 25

one scarf costs $3 more than the other

x = y + 3

now we use that in the first equation again

y+3 + y = 25

2y +3 = 25

2y = 22

y = $11

=> x = y +3 = 11 + 3 = $14

8 0
3 years ago
Solve the following expression when<br> m = 5 and d = 4<br> m2 + 2d + 8 + d
docker41 [41]
M2 + 2d + 8 + d

5(2) + 2(4) + 8 + 4

10 + 8 + 8 + 4

10 + 16 + 4

26 + 4

30
8 0
3 years ago
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
I need help with this one
timama [110]

Answer:

A, it is the 3rd line in a set of 8. 1/8 equals .125 Multiply 3X.125 and it equals .375

Step-by-step explanation:

5 0
3 years ago
Will mark brainliest. will be reported if only for the points
tia_tia [17]

Answer:

d=~7.6

Step-by-step explanation:

7^2+3^2=d^2

49+9=d^2

58=d^2

Square root of 58=~7.6

d=~7.6

6 0
3 years ago
Read 2 more answers
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