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insens350 [35]
3 years ago
14

2√72 + 5√242 what is the answer for this question please can u tell me the right answer​

Mathematics
1 answer:
Radda [10]3 years ago
6 0

Answer:

67\sqrt{2}

Step-by-step explanation:

Assuming you require to simplify the expression.

Using the rule of radicals

\sqrt{a} × \sqrt{b} ⇔ \sqrt{ab}

Simplifying the radicals

\sqrt{72} = \sqrt{36(2)} = \sqrt{36} × \sqrt{2} = 6\sqrt{2}

\sqrt{242} = \sqrt{121(2)} = \sqrt{121} × \sqrt{2} = 11\sqrt{2}

Then

2\sqrt{72} + 5\sqrt{242}

= 2(6\sqrt{2} ) + 5(11\sqrt{2} )

= 12\sqrt{2} + 55\sqrt{2}

= 67\sqrt{2}

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Victoria is interested in opening her own candy store. She has researched and found that of the 14 stores she investigated, the
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Answer:

99% confidence interval for the average start up cost is [76.575 , 123.425].

Step-by-step explanation:

We are given that of the 14 stores Victoria investigated, the average start up cost is 100 thousand dollars with a standard deviation of 29.1 thousand dollars.

So, the pivotal quantity for 99% confidence interval for the population average start up cost is given by;

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

where, \bar X = sample average start up cost = 100 thousand dollars

             \sigma = sample standard deviation = 29.1 thousand dollars

             n = sample of stores = 14

             \mu = population average start up cost

<em>So, 99% confidence interval for the average start up cost, </em>\mu<em> is ;</em>

P(-3.012 < t_1_3 < 3.012) = 0.99

P(-3.012 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 3.012) = 0.99

P( -3.012 \times {\frac{s}{\sqrt{n} } < {\bar X - \mu} < 3.012 \times {\frac{s}{\sqrt{n} } ) = 0.99

P( \bar X-3.012 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +3.012 \times {\frac{s}{\sqrt{n} } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X-3.012 \times {\frac{s}{\sqrt{n} } , \bar X +3.012 \times {\frac{s}{\sqrt{n} } ]

                                                 = [ 100-3.012 \times {\frac{29.1}{\sqrt{14} } , 100+3.012 \times {\frac{29.1}{\sqrt{14} } ]

                                                 = [76.575 , 123.425]

Therefore, 99% confidence interval for the population average start up cost for a candy store is [76.575 , 123.425].

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