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Bond [772]
3 years ago
10

I have no idea how to do Log problems like this. Can anyone help me out?

Mathematics
2 answers:
UkoKoshka [18]3 years ago
6 0
I'm using the following rules:

Rule 1: log(x*y) = log(x)+log(y)
Rule 2: log(x/y) = log(x)-log(y)
Rule 3: log(x^y) = y*log(x)
Rule 4: sqrt(x) = x^(1/2)

log[ (x^2*y)/(sqrt(z)) ] = log[ x^2*y ] - log[ sqrt(z) ]  .... Use Rule 2
log[ (x^2*y)/(sqrt(z)) ] = log[ x^2 ] + log[ y ] - log[ sqrt(z) ] .... Use Rule 1
log[ (x^2*y)/(sqrt(z)) ] = log[ x^2 ] + log[ y ] - log[ z^(1/2) ] .... Use Rule 4
log[ (x^2*y)/(sqrt(z)) ] = 2*log[ x ] + log[ y ] - (1/2)*log[ z ] .... Use Rule 3
log[ (x^2*y)/(sqrt(z)) ] = 2*a + b - (1/2)*c .... Use Substitution 

So 
log[ (x^2*y)/(sqrt(z)) ]
will break down into
2*log[ x ] + log[ y ] - (1/2)*log[ z ]
which is equivalent to 
2a + b - (1/2)c
when a = log(x), b = log(y), c = log(z)

-----------------------------------------------------

So that's why the answer is choice (4) 2a + b - (1/2)c



Virty [35]3 years ago
4 0
Hello there!


The correct answer is option 4

We need to use the rules of logarithms to expand.


I hope this helps!

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Answer:

Respuesta D

Step-by-step explanation:

Paola afirma: Todo número compuesto par, se puede escribir como la multiplicación de factores primos.

Esta afirmación es cierta, pues es un caso  de la afirmación de que todo número natural mayor que uno se puede escribir como multiplicación de números primos. A este proceso se le llama descomposición en factores primos.

Edwin afirma: Todo número compuesto impar se puede escribir como la suma de dos números primos.

Esta afirmación es falsa. Note que al sumar dos números impares de la forma 2k+1 y 2m+1 para k distinto de m, se obtiene

2k+1+2m+1 = 2(km+1)

Es decir, la suma de dos números impares es siempre par.

Note que a excepción de 2, todo número primo es impar. Para que esta afirmación fuera cierta, necesariamente tendría que pasar que cualquier número impar k se escriba de la forma p+2 donde p es un número primo. Esto es equivalente que para cualquier número impar k, el número k-2 sea primo.

Basta con dar un ejemplo para ver que esto no pasa. Tomemos k=11. En este caso, k-2 = 9, el cuál no es un número primo. Entonces 11 no se puede  descomponer como la suma de dos números primos.

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22/30 in lowest terms
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11/15, you would just divide 22/30 in half and that's as far as it can go.
3 0
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Any help would be great
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Answer:

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3 years ago
A total of 500 voters are randomly selected in a certain precinct and asked whether they plan to vote for the Democratic incumbe
lilavasa [31]

Answer:

The limits are 0.6472 and 0.7528

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

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Of the 500 surveyed, 350 said they would vote for the Democratic incumbent.

This means that n = 500, \pi = \frac{350}{500} = 0.7

99% confidence level

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The limits are 0.6472 and 0.7528

6 0
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