asinus

You have 5 segments of a chain and you want to connect them together into one chain. It costs 5 cents to break a link, and 10 cents to weld it back together. What is the least it would cost to join the 5 segments together?

In a row, I break and weld the second and fourth segments.

It costs \(2\times (5cents+10cents)=30cents\)

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 Electric charge is measured in terms of the coulomb (1 C), although this is a very large and not extremely practical unit of measurement.  For example, the charge in a bolt of lightning is about 15 C.  Express the charge in lightning bolt in

a.millicoulombs

b.kilocoulombs

\(Charge\ Lightning\ Bolt\\ Q_{lightning\ bolt}=15C\times\frac{1000mC}{1C}=\color{blue}15\ 000mC\\ Q_{lightning\ bolt}=15C\times\frac{1kC}{1000C}=\color{blue}0.015kC\)

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Max wirft mit einem Würfel einmal. Er gewinnt, wenn er eine 1 oder 6 würfelt. wie groß ist die Wahrscheinlichkeit dafür?

Ja.  2 von 6.

Die Wahrscheinlichkeit zu gewinnen ist also  1 : 3.

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Where is here?

original question: 924500000=___ x 10^---

\(924500000=\frac{92450000}{10^x}\times 10^x\\ x\in \mathbb N\ |x<10\\ \color{blue}x\in \{1,2,3,4,5,6,7,8,9\}\)

Sample:

\(924500000=\frac{92450000}{10^1}\times 10^1=92450000.0\times 10^{1}\\ 924500000=\frac{92450000}{10^2}\times 10^2=9245000.0\times 10^{2}\\ 924500000=\frac{92450000}{10^3}\times 10^3=924500.0\times 10^{3}\\ 924500000=\frac{92450000}{10^4}\times 10^4=92450.0\times 10^{4}\\ …\\ \color{blue}924500000 =9.245\times 10^{8}\)

All right!

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original question: 924500000=___ x 10^---

\(9.245\times 10^x=924500000\)

x=8

\(924500000={\color{blue}9.245} \times 10^{\color{blue}8}\)

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A lattice point in the  -plane is a point both of whose coordinates are integers (not necessarily positive). How many lattice points lie on the hyperbola ?       \(x^2-y^2=17\)

\(x^2-y^2=17\\ y=\pm\sqrt{x^2-17}\)

\(x\in \mathbb Z\ |\ \{[x^2-17]\}\subset \{squares\}\)

There are only four grid points P.

\(P_1(-9,-8)\\ P_2(-9,\ 8)\\ P_3(9,-8)\\ P_4(9,\ 8)\\\)

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3,2,4,7

Find the standard deviation of the data set.
Round your answer to the nearest hundredth.

Standard deviation of the sample

\(s^2=\frac{\sum (x_i-\overline x)^2}{n-1}\)

\(n=4\\ \overline x=\frac{3+2+4+7}{4}\\ \overline x=4\)

\(s^2=\frac{(3-4)^2+(2-4)^2+(4-4)^2+(7-4)^2}{4-1}\\ s^2=\frac{1+4+0+9}{3}\\ s^2=\frac{14}{3}\\ s=\sqrt{\frac{14}{3}}=2.1602469\)

\(s\approx 2.16\)

Standard deviation of the population

\(s^2=\frac{\sum (x_i-\overline x)^2}{n}\)

\(s^2=\frac{(3-4)^2+(2-4)^2+(4-4)^2+(7-4)^2}{4}\\ s^2=\frac{1+4+0+9}{4}\\ s^2=\frac{14}{4}\\ s=\sqrt{\frac{14}{4}}=1.87083\)

\(s\approx 1.87\)

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Hänge mangels Formeln bzw. ähnlicher Beispiele an dieser Aufgabe fest.

Hier die Lösung.

\(\sum M_{(um\ F_1)}=0\)

\(F_2\cdot 1,5m=F_{brett}\cdot 2,5m+F_{yvonne}\cdot 4m\)

\(F_2=\frac{F_{brett}\cdot 2,5m+F_{yvonne}\cdot 4m}{1,5m}\)

\(F_2=\frac{2000N\cdot 2,5m+600N\cdot 4m}{1,5m}\)

\(F_2=4933,\overline 3N\)

\(\sum M_{(um\ F_2)}=0\)

\(F_1\cdot 1,5m=F_{brett}\cdot 1m+F_{yvonne}\cdot 2,5m\)

\(F_1=\frac{F_{brett}\cdot 1m+F_{yvonne}\cdot 2,5m}{1,5m}\)

\(F_1=\frac{2000N\cdot 1m+600N\cdot 2,5m}{1,5m}\)

\(F_1=2333.\overline 3N\)

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