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olga nikolaevna [1]
3 years ago
15

Anyone, please help me please answer my question because I have to pass this tomorrow morning:(

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

Step-by-step explanation:

If you need help with how I got my answer, you can ask me.

\frac{2yx {}^{ - 4} }{(x {}^{ - 4}y {}^{4}) {}^{3}  \times 2x {}^{ - 1}y {}^{ - 3}    }

\frac{2yx {}^{ - 4} }{x {}^{ - 12}y {}^{12}   \times 2x {}^{ - 1}y {}^{ - 3}  }

\frac{2y x {}^{ - 4}  }{2x {}^{  - 13} y {}^{ - 9} }

= x {}^{9} y {}^{ - 8}

=  \frac{x {}^{9} }{y {}^{8} }

12.

( \frac{2u {}^{  4} }{ - u {}^{2}v {}^{ - 1}   \times 2uv {}^{ - 4} } ) {}^{ - 1}

( \frac{2u {}^{4} \times 1 }{ - 2 {u}^{3}v {}^{ - 5}  } ) {}^{ - 1}

( - u v {}^{ 5} ) {}^{   - 1}

=  - u {}^{ - 1} v {}^{ - 5}  =   - \frac{ 1}{uv {}^{5} }

13.

-   \frac{2m {}^{4}n {}^{ - 1}  }{( - m {}^{2}n {}^{ - 2}) {}^{ - 1}    \times  - nm {}^{ - 3} }

-  \frac{2m {}^{4}n {}^{ - 1}  }{ - m {}^{ - 2} n {}^{2}  \times  - nm {}^{ - 3} }

-  \frac{2m {}^{4} n {}^{ - 1} }{m {}^{ - 5}n {}^{3}  }  =  - 2m {}^{9} n {}^{ - 4}

=   - \frac{2m {}^{9} }{n {}^{4} }

14.

( \frac{x {}^{3}y {}^{ - 4}  }{ -  {x}^{2} y {}^{ - 3} \times yx {}^{0}  } ) {}^{3}

( \frac{x {}^{3}y {}^{ - 4}  }{ -  {x}^{ - 2}y {}^{ - 2}  } ) {}^{3}

( -  {x}^{5} y {}^{ - 2} ) {}^{3}  =  -  x {}^{15} y {}^{ - 6}

-  \frac{x {}^{15} }{ {y}^{6} }

15.

-  \frac{yx {}^{4} \times  -  {y}^{3}  z { }^{ - 4} }{(z {y}^{2} ) {}^{4} }

-  \frac{ - y {}^{4} x {}^{4} z {}^{ - 4} }{ {z}^{4} y {}^{8} }  =  - y {}^{4} x {}^{4} z {}^{ - 8}  =  -  \frac{(xy) {}^{4} }{ {z}^{8} }

16.

\frac{h {}^{7}j  {k}^{4} }{4h {}^{4} }  =  \frac{1}{4} h {}^{3}  =  \frac{h {}^{3} jk {}^{4} }{4}

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The weights of 67 randomly selected axles were found to have a variance of 3.85. Construct the 80% confidence interval for the p
VLD [36.1K]

Answer:

3.13

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 67

Variance = 3.85

We have to find 80% confidence interval for the population variance of the weights.

Degree of freedom = 67 - 1 = 66

Level of significance = 0.2

Chi square critical value for lower tail =

\chi^2_{1-\frac{\alpha}{2}}= 51.770

Chi square critical value for upper tail =

\chi^2_{\frac{\alpha}{2}}= 81.085

80% confidence interval:

\dfrac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2}}} < \sigma^2 < \dfrac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2}}}

Putting values, we get,

=\dfrac{(67-1)3.85}{81.085} < \sigma^2 < \dfrac{(67-1)3.85}{51.770}\\\\=3.13

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

8 0
3 years ago
Please help me with this question (25-5) ÷5+8
Kryger [21]
=20\4 +8
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3 0
2 years ago
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Do this please
Kamila [148]

Answer:

-5≤x≤0

Explanation:

You need to use a symbol that shows that x can be both equal to and greater than -5 and also a sign that displays x as less than and equal to 0.

6 0
3 years ago
Find the LCM of the given numbers 9,27<br><br> you'll get 50 points if you answer
Ann [662]

Answer:

27

Step-by-step explanation:

Alternatively, the lcm of 9 and 27 can be found using the prime factorization of 9 and 27: The prime factorization of 9 is: 3 x 3. The prime factorization of 27 is: 3 x 3 x 3. Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(9,9) = 27.

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A mayoral candidate in a large metropolitan area has hired you to take a poll to determine the proportion of registered voters w
sveticcg [70]

Answer:

The smallest sample size that will produce an interval with these specifications is 601.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

He would like you to report a 95% confidence interval with a margin of error no more than 0.04. What is the smallest sample size that will produce an interval with these specifications?

We have to find n for which M = 0.04.

We dont know the true proportion, so we use \pi = 0.5, which is when the smallest sample size needed will have it's largest value.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.04}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2

n = 600.25

Rounding up

The smallest sample size that will produce an interval with these specifications is 601.

6 0
3 years ago
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