<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%20%5Csqrt%7B2%7D%20%2B%201%20%7D%20%20%2B%20%20%5Cfrac%7B1%7D%7B%20%5Csqr

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3 years ago

t%7B2%7D%20%2B%20%20%5Csqrt%7B3%7D%20%20%7D%20%20%2B%20...%20%2B%20%20%5Cfrac%7B1%7D%7B%20%5Csqrt%7Bn%20%2B%201%7D%20%20%2B%20%20%5Csqrt%7Bn%7D%20%7D%20%20%20%5Cleqslant%2020" id="TexFormula1" title=" \frac{1}{ \sqrt{2} + 1 } + \frac{1}{ \sqrt{2} + \sqrt{3} } + ... + \frac{1}{ \sqrt{n + 1} + \sqrt{n} } \leqslant 20" alt=" \frac{1}{ \sqrt{2} + 1 } + \frac{1}{ \sqrt{2} + \sqrt{3} } + ... + \frac{1}{ \sqrt{n + 1} + \sqrt{n} } \leqslant 20" align="absmiddle" class="latex-formula">
help....................

Mathematics

1 answer:

statuscvo [17]3 years ago

8 0

Answer:

all i see is ..........

Step-by-step explanation:

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A triangle with sides of 14 15 and 29 does it form a right triangle

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5x-2x-1x= A.4x B.2x C.8x What is it

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Step-by-step explanation:

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Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind

Gemiola [76]

Answer:

(a)

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

(b)

$1\ girl = \{GBB, BBG, BGB\}$

$P(1\ girl) = 0.375$

ii.

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

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iii.

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Step-by-step explanation:

Given

$Children = 3$

$B = Boys$

$G = Girls$

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

And the number of elements are:

$n(S) = 8$

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

$1\ girl = \{GBB, BBG, BGB\}$

And the number of the list is;

$n(1\ girl) = 3$

The probability is calculated as;

$P(1\ girl) = \frac{n(1\ girl)}{n(S)}$

$P(1\ girl) = \frac{3}{8}$

$P(1\ girl) = 0.375$

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

And the number of the list is;

$n(Atleast\ 2 \ girls) = 4$

The probability is calculated as;

$P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}$

$P(Atleast\ 2 \ girls) = \frac{4}{8}$

$P(Atleast\ 2 \ girls) = 0.5$

Solving (biii) No girl

From the list of possible elements, we have:

$No\ girl = \{BBB\}$

And the number of the list is;

$n(No\ girl) = 1$

The probability is calculated as;

$P(No\ girl) = \frac{n(No\ girl)}{n(S)}$

$P(No\ girl) = \frac{1}{8}$

$P(No\ girl) = 0.125$

7 0

3 years ago

It's in the attachment URGENT

maxonik [38]

Step-by-step explanation:

$\frac{2a}{a + b - c} = \frac{2b}{b + c - a} = \frac{2c}{a + c - b} = k \\ \\ by \: theorem \: on \: equal \: ratios \\ \\ \frac{2a + 2b + 2c}{a + b - c + b + c - a + a + c - b} = k \\ \\ \frac{2(a + b + c)}{a + b + c} = k \\ \\ \therefore \: k = 2$

5 0

3 years ago