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2 years ago

Find the expression below that is equivalent to 5^11/5^13

Mathematics

2 answers:

Mekhanik [1.2K]2 years ago

4 0

$\dfrac{5^{11}}{5^{13}}$

$= 5^{11-13}$

$= 5^{-2}$

$= \dfrac{1}{5^2}$

$= \dfrac{1}{25}$

zalisa [80]2 years ago

3 0

Hi there!

Answer:
$\frac{ 5^{11} }{5 ^{13} } = 5 ^{11-13} = 5^{-2} = \frac{1}{ 5^{2} } = \frac{1}{25} = 0.04$

Additional notes:
In the first step, I used the algebraic rules for the exponents in a fraction, which says the following:
$\frac{ x^{a} }{ x^{b} } = x^{a-b}$

In the third step, I used the algebraic rules for negative exponents, which says the following:
$x^{-a} = \frac{1}{ x^{a} }$

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3 years ago

Tres multiplicaciones de seis factores:una en la que el resultado sea positivo,otra en la que sea negativo,y una cuyo resultado

Free_Kalibri [48]

Answer:

Vamos a inventa tres multiplicaciones de seis factores en la que el resultado sea positivo, es decir un número mayor que cero, en la otra negativo, es decir, un número menor que cero y por último otra multiplicación que de como resultado el cero "0".

Primero recordemos que:

Los factores son los números que se multiplican.

Seis factores serían 6 números.

Multiplicación con resultado positivo, es decir mayor a 0:

2×3×4×2×1×5=240

Multiplicación con resultado negativo, es decir menor a 0:

-4×2×5×2×1×10= -400

Multiplicación con resultado cero:

2×4×6×7×11×0=0

Step-by-step explanation:

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3 years ago

What is 10 times as many as

Talja [164]

Answer:

To answer this question, more information needs to be disclosed.

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2 years ago

Assume that adults were randomly selected for a poll. They were asked if they "favor or oppose using federal tax dollars to fund

ser-zykov [4K]

Testing the hypothesis, it is found that since the p-value of the test is 0.0042 < 0.01, it can be concluded that the proportion of subjects who respond in favor is different of 0.5.

At the null hypothesis, it is tested if the proportion is of 0.5, that is:

$H_0: p = 0.5$

At the alternative hypothesis, it is tested if the proportion is different of 0.5, that is:

$H_1: p \neq 0.5$

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1 - p)}{n}}}$

In which:

$\overline{p}$ is the sample proportion.
p is the value tested at the null hypothesis.
n is the sample size.

In this problem, the parameters are given by:

$p = 0.5, n = 483 + 398 = 881, \overline{p} = \frac{483}{881} = 0.5482$

The value of the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1 - p)}{n}}}$

$z = \frac{0.5482 - 0.5}{\sqrt{\frac{0.5(0.5)}{881}}}$

$z = 2.86$

Since we have a two-tailed test(test if the proportion is different of a value), the p-value of the test is P(|z| > 2.86), which is 2 multiplied by the p-value of z = -2.86.

Looking at the z-table, z = -2.86 has a p-value of 0.0021.

2(0.0021) = 0.0042

Since the p-value of the test is 0.0042 < 0.01, it can be concluded that the proportion of subjects who respond in favor is different of 0.5.

A similar problem is given at brainly.com/question/24330815

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2 years ago

Verify that P = Ce^t /1 + Ce^t is a one-parameter family of solutions to the differential equation dP dt = P(1 − P).

NemiM [27]

Answer:

See verification below

Step-by-step explanation:

We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that

$1-P=1-\frac{ce^t}{1+ce^t}=\frac{1+ce^t-ce^t}{1+ce^t}=\frac{1}{1+ce^t}$

Now, differentiate to obtain

$\frac{dP}{dt}=(\frac{ce^t}{1+ce^t})'=\frac{(ce^t)'(1+ce^t)-(ce^t)(1+ce^t)'}{(1+ce^t)^2}$

$=\frac{(ce^t)(1+ce^t)-(ce^t)(ce^t)}{(1+ce^t)^2}=\frac{ce^t+ce^{2t}-ce^{2t}}{(1+ce^t)^2}=\frac{ce^t}{(1+ce^t)^2}$

To obtain the required form, extract a factor in both the numerator and denominator:

$\frac{dP}{dt}=\frac{ce^t}{1+ce^t}\frac{1}{1+ce^t}=P(1-P)$

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3 years ago