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irakobra [83]
4 years ago
11

(a) Suppose one house from the city will be selected at random. Use the histogram to estimate the probability that the selected

house is valued at less than $500,000. Show your work.
(b) Suppose a random sample of 40 houses are selected from the city. Estimate the probability that the mean value of the 40 houses is less than $500,000. Show your work.

Mathematics
1 answer:
vodka [1.7K]4 years ago
8 0

Answer:

a.    0.71

b.    0.9863

Step-by-step explanation:

a. From the histogram, the relative frequency of houses with a value less than 500,000 is 0.34 and 0.37

-#The probability can therefore be calculated as:

P(500000)=P(x=0)+P(x=500)\\\\\\\\=0.34+0.37\\\\\\\\=0.71

Hence, the probability of the house value being less than 500,000 is o.71

b.

-From the info provided, we calculate the mean=403 and the standard deviation is 278 The probability that the mean value of a sample of n=40 is less than 500000 can be calculated as below:

P(\bar X

Hence, the probability that the mean value of 40 randomly selected houses is less than 500,000 is 0.9863

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Solution :

The formula for the linearization of a function $f(x)$ at a point $x$ = a is given as

$L(x)=f(a)+(x-a)f'(a)$

Assuming the time is t and the distance travelled is $f(t)$, that makes the speed as $f'(t)$.

So substituting them in the linearization formula,

A. At t = 7 minutes

   f(7) = 2.5 km

    f'(7) = 0.5 kpm

  ∴ $L_7(t)=f(7)+(t-7)f'(7)$

             $=2.5+(t-7)0.5$

              $=2.5+0.5t-3.5$

              $=0.5t-1$

B. At t = 18 minutes

      f(18) = 14.8 km

    f'(18) = 0.8 kpm

  ∴ $L_{18}(t)=f(18)+(t-18)f'(18)$

               $=14.8+(t-18)0.8$

              $=14.8+0.8t-14.4$

              $=0.8t-0.4$

C. Substituting the value of t as 14 in both the linearization to determine the position at 14 minutes, we get

$L_7(14)=0.5(14)-1.0$

          = 7 - 1

          = 6 km

$L_{18}(14)=0.8(14)+0.4$

            = 11.2 + 0.4

            = 11.6 km

D. According to the linearization at 7, the distance travelled between the 7 minutes and 14 minutes is = 6 km - 2.5 km

                                           = 3.5 km

And between the 14 minutes and 18 minutes is = 14.8 km - 6 km

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This is an average speed of 0.5 kpm in the first interval and an average speed of 2.2 kpm.

Now, according to the linearization of 18, the distance travelled between the 7 minutes and the 14 minutes is = 11.6 km - 2.5 km

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And between 14 minutes and 18 minutes is = 14.8 km - 11.6 km

                                                                       = 3.2 km

This gives an average speed of 1.3 kpm in the first interval and 0.8 kpm in the second interval.

Therefore, the second approximation is the better one since the average speed are closer to the actual readings in the second linearization.  

4 0
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Answer:

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Step-by-step explanation:

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Hence, the distance d between the park and the library is

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Thus, the distance D between the park and the football field is

D^2=8\cdot (2+8)\\ \\D^2=8\cdot 10\\ \\D^2=80\\ \\D=\sqrt{80}=4\sqrt{5}\approx 8.9\ miles

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