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Alex_Xolod [135]
3 years ago
6

Simplify the expression. (the quantity x to the one-sixth power end quantity to the power of 3.)

Mathematics
2 answers:
Usimov [2.4K]3 years ago
8 0

Answer:

\sqrt{x}

Step-by-step explanation:

If we have that worded expression, we can convert it into actual mathematical terms.

x to the one sixth power: x^{\frac{1}{6}}

That to the power of 3:

(x^{\frac{1}{6}})^3

Using exponent rules, since we have a power to a power, the powers multiply, so:

x^{\frac{1}{6}\cdot 3}\\\\x^{\frac{3}{6}}\\\\ x^{\frac{1}{2}}

If we have a number to a fraction power, the denominator becomes the “find the (denominator) root of x.

So:

x^{\frac{1}{2}} = \sqrt{x}

Hope this helped!

Llana [10]3 years ago
4 0

Answer:

x 1/2 took the quiz

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Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r
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Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                         P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean weight = \frac{\sum X}{n} = 21.88 ounces

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 0.201 ounces

            n = sample of boxes = 12

            \mu = population mean weight

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.796 < t_1_1 < 1.796) = 0.90  {As the critical value of t at 11 degrees of

                                                  freedom are -1.796 & 1.796 with P = 5%}  

P(-1.796 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.796) = 0.90

P( -1.796 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X-1.796 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.796 \times {\frac{s}{\sqrt{n} } } , \bar X+1.796 \times {\frac{s}{\sqrt{n} } } ]

                                        = [ 21.88-1.796 \times {\frac{0.201}{\sqrt{12} } } , 21.88+1.796 \times {\frac{0.201}{\sqrt{12} } } ]

                                        = [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

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