1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
miskamm [114]
3 years ago
11

I am a number greater than 20 but smaller than 30. I have two decimal places. The digit in the ones place is 1 less than the dig

it in the tens places. The value of the digit 3 is 0.3. The digit in the hundredths place is twice the digit in the tenths place. What number am I ?​
Mathematics
1 answer:
Ugo [173]3 years ago
4 0

Answer:

The answer for this problem is 22.36

You might be interested in
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
Need some help please
Gnoma [55]
You can use this to help you. label the triangle like the picture and use the equations. I hope this helps.

3 0
3 years ago
Read 2 more answers
Factorize:<br>(2a - b)² - (a - 2b)²​
bagirrra123 [75]

Answer:

3(a - b)(a + b)

Step-by-step explanation:

Factorize: (2a - b)² - (a - 2b)²​

  • Different of Perfect a Square rule: a²​ - b²​ = (a + b)(a - b)

(2a - b)² - (a - 2b)²​ = [(2a - b) + (a - 2b)] × [(2a - b) - (a - 2b)]

1. Distribute and Simplify:

Distribute the (+) sign on the first bracket and simplify: [(2a - b) + (a - 2b)] → 2a - b + a - 2b → (3a - 3b)

Distribute the (-) sign on the first bracket and simplify: [(2a - b) - (a - 2b)] → 2a - b – a + 2b → (a + b)

We now have:

(3a - 3b)(a + b)

2. Factor out the Greatest Common Factor (3) from 3a - 3b:

(3a - 3b) → 3(a - b)

3. Add "(a + b)" back into your factored expression:

3(a - b)(a + b)

Hope this helps!

7 0
2 years ago
Read 2 more answers
Three components are randomly sampled, one at a time, from a large lot. As each component is selected, it is tested. If it passe
satela [25.4K]

Answer:

The probability of SFS and SSF are same, i.e. P (SFS) = P (SSF) = 0.1311.

Step-by-step explanation:

The probability of a component passing the test is, P (S) = 0.79.

The probability that a component fails the test is, P (F) = 1 - 0.79 = 0.21.

Three components are sampled.

Compute the probability of the test result as SFS as follows:

P (SFS) = P (S) × P (F) × P (S)

            =0.79\times0.21\times0.79\\=0.131061\\\approx0.1311

Compute the probability of the test result as SSF as follows:

P (SSF) = P (S) × P (S) × P (F)

            =0.79\times0.79\times0.21\\=0.131061\\\approx0.1311

Thus, the probability of SFS and SSF are same, i.e. P (SFS) = P (SSF) = 0.1311.

7 0
3 years ago
What's the answer to this?
sleet_krkn [62]
X for left(blue)=43
x for right(red)=17.5
5 0
3 years ago
Other questions:
  • What is the true solution to the logarithmic equation?<br>log2 [log2(sqrt4x)]=1​
    12·2 answers
  • What is the Y intercept of the function Y=(x+2) (x-4)
    11·2 answers
  • PLZ answer with shown proof answer number 9 please
    7·1 answer
  • If the outside temperature was 72 yesterday and today is 16 degrees cooler what is today's temperature
    12·2 answers
  • Ok I need help please
    10·1 answer
  • Which equation can be used to solve for angle A?
    7·1 answer
  • Task 5:
    14·1 answer
  • Show work and simplify the equation
    6·1 answer
  • 1. A soccer field has an area (A = LW) that is represented by the polynomial 1 point
    13·1 answer
  • Pls help me! i’m really confuse on how to tackle this problem (no links)
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!