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viktelen [127]
4 years ago
9

If f(x) = 3x + 2, what is f (5)​

Mathematics
1 answer:
tensa zangetsu [6.8K]4 years ago
8 0
Plug it in into f(x)= 3x+2
Basically
F(5)= (3)(5)+2
= 15+2
=17
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Log14/3 +log11/5-log22/15=log
Tamiku [17]

Answer:

log7

Step-by-step explanation:

when adding logs, apply the log rule: \log _a\left(x\right)+\log _a\left(y\right)=\log _a\left(xy\right)

∴  \log\left(\frac{14}{3}\right)+\log\left(\frac{11}{5}\right)=\log\left(\frac{14}{3}\cdot \frac{11}{5}\right)

when subtracting logs, apply the log rule: \log _a\left(x\right)\:-\:\log _a\left(y\right)=\log _a\left(\frac{x}{y}\right)

\log\left(\frac{14}{3}\cdot \frac{11}{5}\right)-\log\left(\frac{22}{15}\right)=\log\left(\frac{\frac{14}{3}\cdot \frac{11}{5}}{\frac{22}{15}}\right)\\\\=\log\left(7\right)

5 0
2 years ago
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Complement and supplement of a 19
Tems11 [23]

well first the complement I 90-19=71

and supplement is 180-19=161

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
The expression means 15 minus the product of 5 times the difference of p minus 6. If p = 8, then the value of the expression is
MA_775_DIABLO [31]

Answer:

5

Step-by-step explanation:

from the question:

15-5(p-6)

when p=8,

=15-5(8-6)

=15-5(2)

=15-10

=5

please like and Mark as brainliest

3 0
3 years ago
PLZ HELPPPP
Zolol [24]

Answer:

Step-by-step explanation:

3x-4<8

3x<12

x<4

2x+2>4

2x>2

>1

6 0
3 years ago
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