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baherus [9]
3 years ago
10

Geometric mean of 16 and 30

Mathematics
2 answers:
Nataly_w [17]3 years ago
7 0

Answer:

The answer is 21.91

Brut [27]3 years ago
4 0
The geometric mean = √( 3 * 16 ) = √48. = 6.92.
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Jerry starts a savings account with his birthday money of $200. He deposites 10 each week into his account. What is the rate of
MAVERICK [17]

Answer:

C. 10

Step-by-step explanation:

it changes by positive ten each week

3 0
3 years ago
Read 2 more answers
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 number 4 and number 3 please
dangina [55]
For Question 3, 
15      ?
__ = ____
120     100

120?=1500
1500/120=12.5

?=12.5%

Question 4

I think it is $462.80


5 0
3 years ago
Write the equation of the line in point slope form for a horizontal line that passes through (5,-6)
denis-greek [22]

Answer:

y=-6

Step-by-step explanation:

y+6=0(x-5)

Solve

y=-6

7 0
2 years ago
How many 5-member chess teams can be chosen from 15 interested players? Consider only the members selected, not their board posi
ryzh [129]

<u>Answer</u>:

3003 number of 5-member chess teams can be chosen from 15 interested players.

<u>Step-by-step explanation:</u>

Given:

Number of the interested players =  15

To Find:

Number of 5-member chess teams that can be chosen = ?

Solution:

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formulanCr = \frac{n!}{r!(n - r)!}

where

n represents the total number of items,

r represents the number of items being chosen at a time.

Now  we have n = 15 and r = 5

Substituting the values,

15C_5 = \frac{15!}{5!(15- 5)!}

15C_5 = \frac{15!}{5!(10)!}

15C_5 = \frac{15!}{5!(10)!}

15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 }{5 \times 4 \times 3 \times 2 \times 1(10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}

15C_5 = \frac{15\times \times 14 \times 13 \times 12 \times 11}{(5 \times 4 \times 3 \times 2 \times 1)}

15C_5 = \frac{360360}{120}

15C_5 = 3003

7 0
3 years ago
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