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Umnica [9.8K]
2 years ago
14

The population of a town was 25,000 people. then, 3,200 people moved away.. what was the percent decrease?

Mathematics
2 answers:
bogdanovich [222]2 years ago
3 0
It Is B.21.8%, Simple way to figure this out is round B to 20%, If 3200 equals 20% and 20% times 5 equals 100% all you do is multiply 3200 by 5.
puteri [66]2 years ago
3 0
To solve this problem, all you have to do is divide 3,200 by 25,000 and then multiply the number you get by 100 because the question asks for the percentage. Your answer is A. 12.8%
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M/PF Research, Inc. lists the average monthly apartment rent in some of the most expensive apartment rental locations in the Uni
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Answer:

(a) $1,020 or more = 0.2358

(b) Between $880 and $1,130 = 0.7389

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(d) Less than $770 = 0.0294

Step-by-step explanation:

We are given that According to M/PF Research, Inc. report, the average cost of renting an apartment in Minneapolis is $951.

Suppose that the standard deviation of the cost of renting an apartment in Minneapolis is $96 and that apartment rents in Minneapolis are normally distributed.

<em>Let X = apartment rents in Minneapolis</em>

So, X ~ Normal(\mu=$951,\sigma^{2} =$96^{2})

The z score probability distribution for normal distribution is given by;

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

where,  \mu = average cost of renting an apartment = $951

            \sigma = standard deviation = $96

(a) Probability that the price is $1,020 or more is given by = P(X \geq $1,020)

    P(X \geq $1,020) = P( \frac{X-\mu}{\sigma} \geq \frac{1,020-951}{96} ) = P(Z \geq 0.72) = 1 - P(Z < 0.72)

                                                               = 1 - 0.76424 = 0.2358

<em>The above probability is calculated by looking at the value of x = 0.72 in the z table which gives an area of 0.76424.</em>

<em />

(b) Probability that the price is between $880 and $1,130 is given by = P($880 < X < $1,130) = P(X < $1,130) - P(X \leq 880)

    P(X < $1,130) = P( \frac{X-\mu}{\sigma} < \frac{1,130-951}{96} ) = P(Z < 1.86) = 0.96856

     P(X \leq $880) = P( \frac{X-\mu}{\sigma} \leq \frac{880-951}{96} ) = P(Z \leq -0.74) = 1 - P(Z < 0.74)

                                                         = 1 - 0.77035 = 0.22965

<em>The above probability is calculated by looking at the value of x = 1.86 and x = 0.74 in the z table which gives an area of 0.96856 and 0.77035 respectively.</em>

<em />

Therefore, P($880 < X < $1,130) = 0.96856 - 0.22965 = 0.7389

<em />

(c) Probability that the price is between $830 and $940 is given by = P($830 < X < $940) = P(X < $940) - P(X \leq 830)

    P(X < $940) = P( \frac{X-\mu}{\sigma} < \frac{940-951}{96} ) = P(Z < -0.11) = 1 - P(Z \leq 0.11)

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<em>The above probability is calculated by looking at the value of x = 0.11 and x = 1.26 in the z table which gives an area of 0.5438 and 0.89617 respectively.</em>

<em />

Therefore, P($830 < X < $940) = 0.4562 - 0.10383 = 0.3524

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    P(X < $770) = P( \frac{X-\mu}{\sigma} < \frac{770-951}{96} ) = P(Z < -1.89) = 1 - P(Z \leq 1.89)

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