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slava [35]
3 years ago
15

Simplify the following 3√28-5√63+4√112

Mathematics
2 answers:
alexandr402 [8]3 years ago
7 0

7√7

using the ' rule of radicals '

• √a × √b ⇔ √ab

simplifying the radicals

√28 = √(4 × 7 ) = √4 × √7 = 2√7

√63 = √(9 × 7) =√9 × √7 = 3√7

√112 = √(16 × 7 ) = √16 × √7 = 4√7

substituting into the expression

3(2√7) - 5(3√7) + 4(4√7) = 6√7 - 15√7 + 16√7 = 7√7


Dvinal [7]3 years ago
7 0

First, simply each term.  Then, add the coefficients of the like terms (same radicals).

3\sqrt{28} = 3\sqrt{2*2*7} = 3(2)\sqrt{7} = 6\sqrt{7}

5\sqrt{63} = 5\sqrt{3*3*7} = 5(3)\sqrt{7} = 15\sqrt{7}

4\sqrt{112} = 4\sqrt{2*2*2*2*7} = 4(2*2)\sqrt{7} = 16\sqrt{7}

*********************************************************

3\sqrt{28} - 5\sqrt{63} + 4\sqrt{112}

= 6\sqrt{7} - 15\sqrt{7} + 16\sqrt{7}

= (6 - 15 + 16)\sqrt{7}

= 7\sqrt{7}

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Answer:

Step-by-step explanation:

Hello!

To test if boys are better in math classes than girls two random samples were taken:

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Sample 2

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To estimate per CI the difference between the mean percentage that boys obtained in calculus and the mean percentage that girls obtained in calculus, you need that both variables of interest come from normal populations.

To be able to use a pooled variance t-test you have to also assume that the population variances, although unknown, are equal.

Then you can calculate the interval as:

[(X[bar]_1-X[bar_2) ± t_{n_1+n_2-2;1-\alpha /2} * Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }]

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t_{n_1+n_2-2;1-\alpha /2}= t_{15+12-2;1-0.05}= t_{25;0.95}= 1.708

[(82.3-81.2) ± 1.708* (6.11*\sqrt{\frac{1}{15}+\frac{1}{12}  }]

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Using a 90% confidence level you'd expect the interval [-2.94; 5.14] to contain the true value of the difference between the average percentage obtained in calculus by boys and the average percentage obtained in calculus by girls.

I hope this helps!

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