<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B52%7D%20" id="TexFormula1" title=" \sqrt{52} " alt=" \sqrt{52} " align="absmiddle

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Nonamiya [84]

2 years ago

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Mathematics

2 answers:

Papessa [141]2 years ago

6 0

Answer:

i dont get the question

Step-by-step explanation:

Sergio [31]2 years ago

6 0

Answer:

$2\sqrt{13}$

Step-by-step explanation:

Square root of 52

_________________

When finding square root, we always have to find one number that multiplies by itself twice to get the number in the square root.

No number does this for 52, so we have to use the GCF prime, which is :

$\sqrt{2^2} \sqrt{13}$

Apply the radical rule for $\sqrt{2^2}$ :

$\sqrt{2^2} =2$

Therefore :

$2\sqrt{13}$

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umka21 [38]

Answer:

<h3>The answer is 7/12.</h3>

Step-by-step explanation:

$( \frac{2}{3} - \frac{1}{4} ) + ( \frac{1}{2} \times \frac{1}{3} )$

Solve the terms in the bracket first

That's find the LCM of the ones which are subtracting

LCM of 3 and 4 is 12

That's

$\frac{2}{3} - \frac{1}{4} = \frac{8 - 3}{12} = \frac{5}{12}$

Next multiply the other terms in the other bracket

That's

$\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$

Add both results

That's

$\frac{5}{12} + \frac{1}{6}$

Find the LCM

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We get the final answer as

$\frac{5}{12} + \frac{1}{6} = \frac{5 + 2}{12} = \frac{7}{12}$

Hope this helps you

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3 years ago

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Vsevolod [243]

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3 years ago

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Answer:

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Step-by-step explanation:

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Answer: $P(28.5$ And we can find this probability with this difference:

$P(-0.7$

And in order to find these probabilities we use tables for the normal standard distribution, excel or a calculator.

$P(-0.7$

So then the best answer would be 21.82 %

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

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