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HACTEHA [7]
3 years ago
12

Simplify (4xy^-2)/(12x^(-1/3)y^-5) and Show work ...?

Mathematics
1 answer:
oee [108]3 years ago
8 0
The answer is \frac{x^{4/3}y^{3}}{3}.

The fraction is: \frac{4xy^{-2} }{12 x^{-1/3} y^{-5}  }
Let's rewrite it: \frac{4xy^{-2} }{12 x^{-1/3} y^{-5} }=\frac{4 }{12 }*\frac{x }{ x^{-1/3} }*\frac{y^{-2} }{ y^{-5} }

Since \frac{x^{a} }{x^{b}} =x^{a-b},
then:
*** \frac{x}{ x^{-1/3} }= x^{1-(-1/3)} = x^{3/3+1/3} = x^{4/3}
*** \frac{y^{-2} }{ y^{-5} }=  y^{-2-(-5)} = y^{-2+5} = y^{3}
______
\frac{4 }{12 }*\frac{x }{ x^{-1/3} }*\frac{y^{-2} }{ y^{-5} }= \frac{1*4}{3*4}* x^{4/3}*y^{3}= \frac{1}{3} * x^{4/3}*y^{3}= \frac{x^{4/3}y^{3}}{3}
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Hi guys can you answer my math question​
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Step-by-step explanation:

Put 1 on the far left and 4 on the far right.

Put. 2 the top and 3 at the bottom.

Put 5 in the centre.

There are 2 ways you can find to to make all the numbers add up to 5.

The number in the centre is the common anwser to the sums.

4 0
3 years ago
£132 in the ratio 3 : 8
Vedmedyk [2.9K]

Answer:

£36 : £96

Step-by-step explanation:

3 + 8 = 11

132 ÷ 11 = 12

⇒ 12 x 3 : 12 x 8

⇒ £36 : £96

5 0
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3 years ago
Is 4/6 more than. 4/8
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4 0
3 years ago
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A random sample of 21 observations is used to estimate the population mean. The sample mean and the sample standard deviation ar
stepladder [879]

Answer:

a. CI=[128.79,146.41]

b. CI=[122.81,152.39]

c. As the confidence level increases, the interval becomes wider.

Step-by-step explanation:

a. -Given the sample mean is 137.6 and the standard deviation is 20.60.

-The confidence intervals can be constructed using the formula;

\bar X\pm \ z\frac{s}{\sqrt{n}},

where:

  • s is the sample standard deviation
  • z is the s value of the desired confidence interval

we then calculate our confidence interval as:

\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.05/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm1.960\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm8.8108\\\\\\=[128.789,146.411]

Hence, the 95% confidence interval is between 128.79 and 146.41

b. -Given the sample mean is 137.6 and the standard deviation is 20.60.

-The confidence intervals can be constructed using the formula in a above;

\bar X\pm z\frac{s}{\sqrt{n}}\\\\=137.60\pm z_{0.01/2}\times\frac{20.60}{\sqrt{21}}\\\\=137.60\pm3.291\times \frac{20.60}{\sqrt{21}}\\\\=137.60\pm 14.7940\\\\\\=[122.806,152.394]

Hence, the variable's 99% confidence interval is between 122.81 and 152.39

c. -Increasing the confidence has an increasing effect on the margin of error.

-Since, the sample size is particularly small, a wider confidence interval is necessary to increase the margin of error.

-The 99% Confidence interval is the most appropriate to use in such a case.

7 0
4 years ago
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