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saul85 [17]
3 years ago
10

Simplify completely: Show all work for full credit.

Mathematics
1 answer:
solong [7]3 years ago
6 0

Answer:

3 {x}^{3}

Step-by-step explanation:

( \frac{81 {x}^{12} }{3 {x}^{3} }  {)}^{ \frac{1}{3} }

( \frac{27 {x}^{12} }{ {x}^{3} }  {)}^{ \frac{1}{3} }  = (27 {x}^{9}  {)}^{ \frac{1}{3} }

\sqrt[3]{27 {x}^{9} }

= 3 {x}^{3}

You might be interested in
What is the conjugate? √x + 2√b
Llana [10]
The correct answer is:  " √x  −  <span>2√b " .
</span>_________________________________________________________

       The "conjugate" of  " √x  + 2√b "  is:  " √x  −  2√b " .
_________________________________________________________

Explanation:
_________________________________________________________
      In an expression with 2 (TWO) terms;  that is, in a "binomial expression", 
the "conjugate" of that expression refers to that very expression — with the "sign" in between those two terms—"reverse" (e.g. "minus" becomes "plus" ; or, "plus" becomes "minus" .) .
_________________________________________________________ 

      →  So:  We are given:  " <span>√x + 2√b " .
</span>
      →  Note that this is a "binomial expression" ;

      →  that is, there are 2 (TWO) terms:  " <span>√x " ; and:  " 2√b " .
_________________________________________________________

To find the "conjugate" of the given binomial expression:

     </span>→  " <span>√x + 2√b "  ; 

     </span>→  We simply change the "+" {plus sign} to a "<span>−"  {minus sign} ;  and rewrite:
___________________________________________________________

     </span>→  " √x − 2√b "  ; 

     →  which is the "conjugate" ; and is the correct answer:
___________________________________________________________
→  " √x − 2√b " ;   is the "conjugate" of the expression:  " <span>√x + 2√b " . 
___________________________________________________________
     
</span>→  {that is:  " √x − 2√b " ;  is the conjugate.}.
___________________________________________________________
7 0
3 years ago
Assuming that the population is normally​ distributed, construct a 9090​% confidence interval for the population mean for each o
jasenka [17]

Given:

Set A: 1 4 4 4 5 5 5 8

Mean: 4.5

Standard dev: 1.9

 

Set B:

Mean: 4.5

Standard dev: 2.45

 

% =  90 

Set A:

Standard Error, SE = s/ √n =    1.9/√8 = 0.67  

Degrees of freedom = n - 1 =   8 -1 =  7   

t- score =  1.89457861

<span> <span><span> <span>   </span> </span> </span></span>

Width of the confidence interval = t * SE =     1.89457861* 0.67 = 1.272685913

Lower Limit of the confidence interval = x-bar - width =      4.5 - 1.272685913 = 3.23

Upper Limit of the confidence interval = x-bar + width =      4.5 + 1.272685913 = 5.77

The 90% confidence interval is [3.23, 5.77]

 

Set B:

Standard Error, SE = s/ √n =    2.45/√8 = 0.87  

Degrees of freedom = n - 1 =   8 -1 = 7   

t- Score =  1.89457861

<span> <span><span> <span>   </span> </span> </span></span>

Width of the confidence interval = t * SE =     1.89457861* 0.87 = 1.641094994

Lower Limit of the confidence interval = x-bar - width =      4.5 - 1.641094994 = 2.86

Upper Limit of the confidence interval = x-bar + width =      4.5 + 1.641094994 = 6.14

The 90% confidence interval is [2.86, 6.14]

 

<span>We can obviously see that sample B has more variation in the scores than sample A. The fact that the standard deviation is 2.45 for B and 1.9 for A). Therefore, they yield dissimilar confidence intervals even though they have the same mean and range.</span>

6 0
3 years ago
What times blank equal 90 and a 100
torisob [31]

Answer:

10x9=90

10x10=100

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10
jasenka [17]

<u>Answer:</u>

 The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

<u>Solution: </u>

The given expression is (2 x-3 y)^{10}

As per binomial theorem, we know,

(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now x^{5}\times y^5 will be the 6 term where k =5

Now, \mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}

So, the coefficient of x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right.

10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252

The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

6 0
3 years ago
Pls help me and thank ya!!
algol13

Answer:

Option 1

Step-by-step explanation:

Rational numbers do not have infinite decimals.

Option 1:

-5, 3/4 and √49 are all rational. Correct

√49 = 7

3/4 = 0.75

-5 = -5

Option 2:

1/4 = 0.25

5 = 5

√12 = 3.4641...

Incorrect

Option 3:

-1/2 = -0.5

-3 = -3

√8 = 2.82841...

Incorrect

Option 4:

1/7 = 0.142857...

9 = 9

√11  = 3.31662...

Incorrect

5 0
3 years ago
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