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ehidna [41]
3 years ago
6

In March 2007​, the U.S. unemployment rate was 4.4 percent. In August 2008​, the unemployment rate was 6.1 percent. Predict what

happened to unemployment between March 2007 and August 2008​, if the labor force was constant. If the labor force remained constant between March 2007 and August 2008​, then the number unemployed​ _______.
Mathematics
1 answer:
Luba_88 [7]3 years ago
4 0

Answer:

Increased.

Step-by-step explanation:

In March 2007, the unemployment rate was 4.4 percent. In August 2008 was 6.1 percent. We need to remember that the unemployment rate equals the number of unemployed people divided by the people in the labor force.

Now, if we consider that the labor force remained constant during this period of time (according to the problem) then this would mean that the number of unemployed people actually increased during this period of 17 months.

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Altogether, Alicia has to drive 104 miles on the turnpike.
omeli [17]
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6 0
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(3/1/2 - 9/3/4) / (-2.5)=
Alex73 [517]
<h3>Answer :</h3>

\:  \:  \:

=  \rm \large  \frac{3 \frac{1}{2}  - 9 \frac{3}{4} }{ - 2.5}

\:  \:

= \rm \large  \frac{ \frac{7}{2}  - 9 \frac{3}{4} }{ - 2.5}

\:  \:

=  \rm \large \:  \frac{ \frac{7}{2} -  \frac{39}{4}  }{ - 2.5}

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=  \rm \large \:  \frac{ \frac{25}{4} }{ - 2.5}

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6 0
3 years ago
What are the zeros of the function? what are their multiplicities?<br> f(x)=x^4-4x^3+3x^2
-Dominant- [34]
Since each term contains x, x= 0 is one answer Also as x^2 is a factor  it has duplicity 2.
Factoring:-

x^2 (x^2 - 4x + 3) = 0
x^2( x - 3)(x - 1) = 0

so x = 3 and x = 1 are also zeros


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4 0
3 years ago
You work with the office of city planning which is currently evaluating a new contract for a major highway. The chosen contracto
Ipatiy [6.2K]

Answer:

0.5488 = 54.88% probability that there are no defects in the completed highway.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

The portion of highway that your city is building is 30 miles long.

The mean is one defect per 50 miles. Since the highway is 30 miles, we have that:

\mu = \frac{30}{50} = 0.6

What is the probability that there are no (0) defects in the completed highway?

This is P(X = 0). So

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.6}*0.6^{0}}{(0)!} = 0.5488

0.5488 = 54.88% probability that there are no defects in the completed highway.

7 0
3 years ago
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