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RoseWind [281]
3 years ago
15

The number of private investigators is expected to increase from 52,000 in 2006 to 61,000 in 2016. If there are 648,982 police o

fficers in 2006, how many police officers must there be in 2016 to have a greater percent increase than that of private investigators?
a.
657,983
b.
761,307
c.
671,128
d.
713,180
Mathematics
2 answers:
katovenus [111]3 years ago
7 0

Answer:

Option: B is correct.

( b.  761,307 )

Step-by-step explanation:

It is given that:

The number of private investigators is expected to increase from 52,000 in 2006 to 61,000 in 2016.

i.e. the percentage increase is given by:

\dfrac{61000-52000}{52000}\times 100\\\\=\dfrac{9000}{52000}\times 100\\\\=17.3\%

Now it is given that:

There are 648,982 police officers in 2006.

so the number of police officers in 2016 will be:

117.3\% \times 648982\\\\=\dfrac{117.3}{100}\times 648982\\\\=761255.886

Hence, number of  police officers  in 2016 to have a greater percent increase than that of private investigators is:

Option:b 761,307

(Since in all the other option the value is less than the estimated value i.e. 761255.886 )

finlep [7]3 years ago
4 0
<span>b. 761,307 ................</span>
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Delvig [45]

Answer:

The correct answer is - 768.

Step-by-step explanation:

The given question is based on a specific pattern present in between the numbers in the given series -

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let us find the relation between the initial two number:

25 - 50, here number 50 is two times of 25 or (x*2-0)

relation between 2nd and 3rd number:

50 - 99, here 99 is two times of 50 minus 1 or (x*2-1)

relation between 3rd and 4th number:

99 - 196, here 196 is two times of 99 minus 2 or (x*2-2)

relation between 4th and 5th number:

196 - 388, here 388 is the two times of 196 minus 4 or (x*2-4)

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5 0
2 years ago
Mathematical x4 - x2 ≥ 2x
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4 0
3 years ago
Exhibit 6-5 The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation o
hjlf

Answer:

46.18% of the items will weigh between 6.4 and 8.9 ounces.

Step-by-step explanation:

We are given that the weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces.

<em>Let X =  weight of items produced by a machine</em>

The z-score probability distribution for is given by;

                Z = \frac{  X -\mu}{\sigma}  ~ N(0,1)

where, \mu = mean weight = 8 ounces

            \sigma = standard deviation = 2 ounces

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, percentage of items that will weigh between 6.4 and 8.9 ounces is given by = P(6.4 < X < 8.9) = P(X < 8.9 ounces) - P(X \leq 6.4 ounces)

   P(X < 8.9) = P( \frac{  X -\mu}{\sigma} < \frac{  8.9-8}{2} ) = P(Z < 0.45) = 0.67364  {using z table}

   P(X \leq 70) = P( \frac{  X -\mu}{\sigma} \leq \frac{  6.4-8}{2} ) = P(Z \leq -0.80) = 1 - P(Z < 0.80)

                                                 = 1 - 0.78814 = 0.21186

<em>Now, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.45 and x = 0.80 in the z table which has an area of </em>0.67364<em> and </em>0.78814<em> respectively.</em>

Therefore, P(6.4 < X < 8.9) = 0.67364 - 0.21186 = 0.4618 or 46.18%

<em>Hence, 46.18% of the items will weigh between 6.4 and 8.9 ounces.</em>

8 0
3 years ago
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