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Stells [14]
3 years ago
5

Simplify the - 501/2+12.3

Mathematics
1 answer:
SOVA2 [1]3 years ago
4 0

Answer:

- 50 \frac{1}{2}  + 12.3 \\  \\  =  \frac{ - 101}{2}  + 12.3 \\  \\  =  - 38.2 \\  \\  =  - 38 \frac{1}{5}

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1. The sum of two number is x. If one of the numbers is 16, what is the other number?
Vlada [557]
I hope this helps you




first number 16



second number ?



Sum of these numbers =x


x=16+?


?=x-16



3 0
3 years ago
What number should be placed in the box to help complete the division calculation? Long division setup showing an incomplete cal
lianna [129]

Answer:

The unknown value is being subtracted from 226 is 160

Step-by-step explanation:

Long division setup showing an incomplete calculation

  • 16 is in the divisor
  • 3426 is in the dividend
  • 2 hundreds and 1 tens is written in the quotient
  • 3200 is subtracted from 3426 to give 226
  • An unknown value represented by a box is being subtracted from 226

so,

The dividend = 3426

 The divisor = 16

2 hundreds means 200 and 1 tens means 10

∵ The quotient = 200 + 10 + x

∵ Dividend = divisor × quotient

∴ (16 × 200) + (16 × 10) + (16 × x) +remainder  = 3426  

∵ 16 × 200 = 3200

Subtract 3200 from the dividend  

∴ 3426 - 3200 = 226

∵ 16 × 10 = 160

∴ 226- 160 = 66

⇒160is the unknown value

∵ 16 × x = 16x  

∵ 66 - 16x = 0

∴ 66 = 16x

- Divide both sides by 16

∴ x = 4  and remainder = 2

∴ 3426 ÷ 16 = 200 + 10 + 4

∴ 3426 ÷ 16 = 214  

∴ From the steps above the missing number subtracted from

  226 is 160

The unknown value is being subtracted from 226 is 160

3 0
2 years ago
PLEASE HELP SOLVE FOR ‘X’!!!
antoniya [11.8K]

Answer:

x= 10

Step-by-step explanation:

1/5 x -2/3 = 4/3

Add 2/3 to each side

1/5x -2/3 +2/3 = 4/3 +2/3

1/5x = 6/3

1/5x = 2

Multiply each side by 5

1/5x * 5 = 2*5

x = 10

3 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
Help whats the answer​
Gnom [1K]

Answer:36cm

Step-by-step explanation:

8 0
3 years ago
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