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borishaifa [10]
3 years ago
12

Log3 a/3 = ??????????????????????????

Mathematics
2 answers:
Naddik [55]3 years ago
8 0
We have that
<span>Log3 a/3
</span>Rewrite log3(a/3) using the change of base <span>formula

we know that
</span>The change of base rule can be used if a and b are greater than 0 and not equal to 1, and x is greater than 0<span>.
</span>so
loga(x)=<span>logb(x)/<span>logb<span>(a)
</span></span></span>Substitute in values for the variables in the change of base <span>formula
</span>
in this problem
b=10
a=3
x=a/3

log3(a/3)=[log (a/3)]/[log (3)]

the answer is
[log (a/3)]/[log (3)]

Elis [28]3 years ago
3 0

Answer:

-1.631

Step-by-step explanation:

Got it on edge :)

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[20points][Precal] Find the cross product of -(3/4)V and (1/2)w if v = &lt;-2, 12, -3&gt; and w = &lt;-7, 4, -6&gt;
Lana71 [14]
\mathbf v=\langle-2,12,-3\rangle=-2\,\vec i+12\,\vec j-3\,\vec k
-\dfrac34\mathbf v=\dfrac32\,\vec i-9\,\vec j+\dfrac94\,\vec k

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\dfrac12\mathbf w=-\dfrac72\,\vec i+2\,\vec j-3\,\vec k

-\dfrac34\mathbf v\times\dfrac12\mathbf w=\begin{vmatrix}\vec i&\vec j&\vec k\\\frac32&-9&\frac94\\-\frac72&2&-3\end{vmatrix}
=\left((-9)\times(-3)-\dfrac94\times2\right)\,\vec i-\left(\dfrac32\times(-3)-\dfrac94\times\left(-\dfrac72\right)\right)\,\vec j+\left(\dfrac32\times2-(-9)\times\left(-\dfrac72\right)\right)\,\vec k
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6 0
3 years ago
You are given the following information obtained from a random sample of 5 observations. 20 18 17 22 18 At 90% confidence, you w
Margaret [11]

Answer:

a

  The null hypothesis is  

         H_o  : \mu  =  21

The Alternative  hypothesis is  

           H_a  :  \mu<   21

b

     \sigma_{\= x} =   0.8944

c

   t = -2.236

d

  Yes the  mean population is  significantly less than 21.

Step-by-step explanation:

From the question we are given

           a set of  data  

                               20  18  17  22  18

       The confidence level is 90%

       The  sample  size  is  n =  5  

Generally the mean of the sample  is  mathematically evaluated as

        \= x  =  \frac{20 + 18 +  17 +  22 +  18}{5}

       \= x  =  19

The standard deviation is evaluated as

        \sigma =  \sqrt{ \frac{\sum (x_i - \= x)^2}{n} }

         \sigma =  \sqrt{ \frac{ ( 20- 19 )^2 + ( 18- 19 )^2 +( 17- 19 )^2 +( 22- 19 )^2 +( 18- 19 )^2 }{5} }

         \sigma = 2

Now the confidence level is given as  90 %  hence the level of significance can be evaluated as

         \alpha = 100 - 90

        \alpha = 10%

         \alpha =0.10

Now the null hypothesis is  

         H_o  : \mu  =  21

the Alternative  hypothesis is  

           H_a  :  \mu<   21

The  standard error of mean is mathematically evaluated as

         \sigma_{\= x} =   \frac{\sigma}{ \sqrt{n} }

substituting values

         \sigma_{\= x} =   \frac{2}{ \sqrt{5 } }

        \sigma_{\= x} =   0.8944

The test statistic is  evaluated as  

              t =  \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }

substituting values

              t =  \frac{ 19  - 21 }{ 0.8944 }

              t = -2.236

The  critical value of the level of significance is  obtained from the critical value table for z values as  

                   z_{0.10} =  1.28

Looking at the obtained value we see that z_{0.10} is greater than the test statistics value so the null hypothesis is rejected

6 0
3 years ago
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