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3 years ago

Help me, please!! Estoy desesperada y no hablo inglês!! Help

Mathematics

2 answers:

Aleksandr-060686 [28]3 years ago

8 0

$D:x>0\\
\log x+\log x^2+\log x^3=-6\\
\log (x\cdot x^2\cdot x^3)=\log 10^{-6}\\
x^6=10^{-6}\\
x^6=\left(\frac{1}{10}\right)^6\\ \boxed{x=\frac{1}{10}}$

goblinko [34]3 years ago

7 0

Log x^n = nLog x
log x + log x^2 + log x^3
Log x + 2 log x +3 log x = -6
6 log x = -6
Log x = -1
X=10^-1 = 1/10
Respuesta final
X=1/10

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2 years ago

The amount of time workers spend commuting to their jobs each day in a large metropolitan city has a mean of 70 minutes and a st

Zepler [3.9K]

Answer:

At least 75% of these commuting times are between 30 and 110 minutes

Step-by-step explanation:

Chebyshev Theorem

The Chebyshev Theorem can also be applied to non-normal distribution. It states that:

At least 75% of the measures are within 2 standard deviations of the mean.

At least 89% of the measures are within 3 standard deviations of the mean.

An in general terms, the percentage of measures within k standard deviations of the mean is given by $100(1 - \frac{1}{k^{2}})$ .

In this question:

Mean of 70 minutes, standard deviation of 20 minutes.

Since nothing is known about the distribution, we use Chebyshev's Theorem.

What percentage of these commuting times are between 30 and 110 minutes

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THis means that 30 and 110 minutes is within 2 standard deviations of the mean, which means that at least 75% of these commuting times are between 30 and 110 minutes

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3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

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3 years ago

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