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The site security certificate is out of date. Admin needs to do something about it.

Are other people - maybe on this English forum - having a problem with login from their  FireFox  browser?

Is any who is using a different browser having any problems??

An acre is 4047 square metres

Hi Omi,

How long have you not been able to log in for?  Maybe it is just temporary ??

I can't read German so I am not too sure what the error message says. 

I have just been over to the German forum and I had no trouble logging in. 

I am using Google Chrome.

I will contact Mr Massow, if I can but I want to feel sure that it is a continuing problem first.  
Not one that will go away by itself shortly. ://

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Hallo Omi,

Wie lange warst du nicht in der Lage, zum anmelden?  Vielleicht ist es nur vorübergehend?

Ich kann nicht Deutsch lesen, so bin ich nicht sicher, was die Fehlermeldung sagt.

Ich habe gerade mehr als zum deutschen Forum und ich hatte keine Probleme bei der Anmeldung.

Ich verwende Google Chrome.

Ich werde Herrn Massow, wenden, wenn ich kann, aber ich sicher das Gefühlmöchte, dass es ein ständiges Problem zuerst.

Nicht eine, die von selbst kurz verschwinden wird. ://

y=127/20*x

two dice are rolled.

what is the probability that the total number achieved

is greater than 4 but less than or equal to ten?

\(\color{green}{\text{In green dice one}}\\ \color{blue}{\text{In blue dice two}}\\ \color{red}{\text{In red the total number } >4 \text{ and } \le 10 }\)

\(\begin{array}{|r|r|r|r|r|r|r|} \hline + & \color{blue}1 & \color{blue}2 & \color{blue}3 & \color{blue}4 & \color{blue}5 & \color{blue}6 \\ \hline \color{green}1 & 2 & 3 & 4 & \color{red}5 & \color{red}6 & \color{red}7 \\ \hline \color{green} 2 & 3 & 4 & \color{red}5 & \color{red}6 & \color{red}7 & \color{red}8 \\ \hline \color{green}3 & 4 & \color{red}5 & \color{red}6 & \color{red}7 & \color{red}8 & \color{red}9 \\ \hline \color{green}4 & \color{red}5 & \color{red}6 & \color{red}7 & \color{red}8 & \color{red}9 & \color{red}10 \\ \hline \color{green}5 & \color{red}6 & \color{red}7 & \color{red}8 & \color{red}9 & \color{red}10& 11 \\ \hline \color{green}6 & \color{red}7 & \color{red}8 & \color{red}9 & \color{red}10& 11& 12 \\ \hline \end{array} \)

\(\text{The probability of } > 4 \text{ but } \le 10 \text{ is } \dfrac{27}{36} \text{ or } 75\ \% \)

Prostate cancer probabilities:

.

       Hi Denzel, you are our music man, I remember you well        

how many digits does 2^1224 have

Number of precomma places:

\(\begin{array}{|rcll|} \hline && \lfloor 1 + \log_{10}{(2^{1224})} \rfloor \\ &=& \lfloor 1 + 1224\cdot \log_{10}{(2)} \rfloor \quad & | \quad \log_{10}{(2)} = 0.30102999566 \\ &=& \lfloor 1 + 1224\cdot 0.30102999566 \rfloor \\ &=& \lfloor 1 + 368.460714693 \rfloor \\ &=& \lfloor 369.460714693 \rfloor \\ &=& 369 \\ \hline \end{array}\)

2¹²²⁴ has 369 digits

-3(c+ 4) - 2 > 7

-3c -12 - 2 > 7

-3c -14 > 7

-3c  > 21

c < -7

x(1*234)(85)^-2.75*x^sqrt(34)=66

\(x(1*234)(85)^{-2.75}*x^\sqrt{34}=66\\ 234x*(85)^{-2.75}*x^\sqrt{34}=66\\ 234*(85)^{-2.75}*x^{\sqrt{34}+1}=66\\ x^{\sqrt{34}+1}=\frac{66*(85)^{2.75}}{234}\\ x=\left(\frac{66*(85)^{2.75}}{234}\right)^{\frac{1}{1+\sqrt{34}}}\\ x=\left(\frac{66*(85)^{2.75}}{234}\right)^{\frac{1-\sqrt{34}}{1-34}}\\ x=\left(\frac{66*(85)^{2.75}}{234}\right)^{\frac{\sqrt{34}-1}{33}}\\ \)

((66*85^2.75/234)^((sqrt(34)-1)/33) = 4.969083918033439

Solve for x over the real numbers:
0.00115695 x^(1+sqrt(34)) = 66

Divide both sides by 0.00115695:
x^(1+sqrt(34)) = 57046.6

Raise both sides to the power of 1/(1+sqrt(34)):
Answer: |x = 4.96908

234x(~0.00000494422)(x^34^[1/2])=66

234x(x^17)=13348915.4175

x^18=57046.6470833

x=57046.6470833^(1/18)

x=~1.837532406

Linear regression, in laymans terms, is making a line of best fit when there's only one x variable for points of data.

If the angles are congruent, then

7x+15 = 2x+12

5x = -3

x = -3/5

2^x = 6

Reversed, this equation becomes log with a base of 2 of 6 equals x.

so enter log2(6) into the calculator and you should get 2.58

I don't know who you are, but you just made my day! Thank you for that :)

First of all, -4 * 5 = -20.

Then, n^5 * n^7 = n^(5+7) = n^12

So the final answer is -20n^12

The answer is 235

FV = PV [1 + R]^N

50,000 = 9,000[1 + 10/1200]^N    divide both sides by 9,000

5.5556 = [1 + 0.0083333]^N

5.5556 = [1.0083333]^N                take the log of both sides

N = Log(5.5556) / Log(1.0083333]

N = 206.63 months, or about 17 years and 3 months.

23.5

this cannot be reduced

A success is when the first digit is 1 or 2.