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kondor19780726 [428]
4 years ago
13

Multiply and simplify to lowest terms when appropriate 11/9× 27/55

Mathematics
1 answer:
arsen [322]4 years ago
5 0
11/9 * 27/55
First, you need to transform this expression to one fraction, by multiplying the numerators between each other and put the answer as one numerator, then the denominators and put the answer as one denominator.
11/9 * 27/55 = 297/495.
Now comes the simplifying. You need to divide both numerator and denominator by the same number, till you get an irreducible fraction (a fraction In which numerator and denominator can't be divided by the same number to get another whole number).
In this case, both 297 and 495 are divisible by 99.
297/99 = 3 and 495/99 = 5.
So 297/495 = 3/5.
3 and 5 can't be divided by a same number, so 3/5 represents the irreducible and simplified fraction.

You can re-check your answer (very important):
11/9 * 27/55 = 0.6
297/495 = 0.6
3/5 = 0.6
11/9*22/55 = 297/495 = 3/5
The answer has been approved.

Hope this Helps! :) 
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Add and subtract the rational expressions. 7x+5/x-1 -(8x1/x-1+2x-4/x-1) Which statements are true about the process for adding a
anzhelika [568]

Answer:

2. The denominator of the fully simplified expression will be x – 1.

4. The numerator of the fully simplified expression will be –3x + 10.

Step-by-step explanation:

Given the rational expression

\dfrac{7x+5}{x-1} -(\dfrac{8x-1}{x-1} +\dfrac{2x-4}{x-1} )

Let us first simplify before making our deductions.

Opening the brackets

\dfrac{7x+5}{x-1} -(\dfrac{8x-1}{x-1} +\dfrac{2x-4}{x-1} )=\dfrac{7x+5}{x-1} -\dfrac{8x-1}{x-1} -\dfrac{2x-4}{x-1}

Taking LCM

=\dfrac{7x+5-(8x-1)-(2x-4)}{x-1}

Opening the brackets and simplifying

=\dfrac{7x+5-8x+1-2x+4}{x-1}\\\text{Collecting like terms in the numerator}\\=\dfrac{7x-8x-2x+5+1+4}{x-1}\\=\dfrac{-3x+10}{x-1}

The following statements are therefore true:

2. The denominator of the fully simplified expression will be x – 1.

4. The numerator of the fully simplified expression will be –3x + 10.

5 0
3 years ago
Noise levels at 3 airports were measured in decibels yielding the following data:
Shalnov [3]

Answer:

a)  The 90% confidence interval for the mean noise level at such locations      

  (112.46 , 163.54)

b) The critical value that should be used in constructing the confidence interval

Z₀.₁₀ = 1.645

Step-by-step explanation:

<u><em>Step(i)</em></u> :-

Noise levels at 3 airports were measured in decibels yielding the following data

108   146   160

Mean of given data

         x^{-} = \frac{108+146+160}{3} = 138

x          :     108   146   160

x-x⁻      :      -30   8       22

(x-x⁻ )² :     900   64     484

Variance  σ ² = ∑(x-x⁻ )²/ n-1

                                        = \frac{900+64+484}{3-1}= 724

Standard deviation  

                σ = √724 = 26.90

<u><em>Step(ii):</em></u>-

The 90% confidence interval for the mean noise level at such locations                        

       (x^{-} - Z_{0.90} \frac{S.D}{\sqrt{n} } , (x^{-} + Z_{0.90} \frac{S.D}{\sqrt{n} } )

The critical value that should be used in constructing the confidence interval

Z₀.₁₀ = 1.645

         (138 - 1.645 \frac{26.90}{\sqrt{3} } , (138 + 1.645 \frac{26.90}{\sqrt{3} } )

        ( 138 - 25.54 , 138 + 25.54 )

        (112.46 , 163.54)

4 0
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Answer:

a

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10 = 2x ( divide both sides by 2 )

5 = x or x = 5

8 0
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