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1 year ago

Evaluate b(x)=18−0.5x when x=−2,0, and 5.

Mathematics

1 answer:

Savatey [412]1 year ago

5 0

hope that helps :))

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1) A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2,

crimeas [40]

Answer:

Step-by-step explanation:

Let us record the station number 1, 2 or 3 for each family member A, B or C.

I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.

From table it is clear that the total outcomes possible are 27.

We know that, probability can be defined as,

$PROBABITILY = \frac{NUMBER\;OF\;DESIRED\;OUTCOMES}{TOTAL\;NUMBER\;OF\;OUTCOMES}$

a) All Members Assigned to the Same Station.

Cases for all members being assigned to same station are as follows:

[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).

Therefore,

$PROBABILITY\;(Case\;a) = \frac{3}{27}\\\\PROBABILITY\;(Case\;a) = 0.111$

b) At Most Two Members Assigned to the Same Station.

It means that maximum of 2 members can have the same station. Cases for this situation are as follows:

[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],

[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]

(outcome number 2, 3, 4, 5, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 25 and 26 in the table).

Therefore,

$PROBABILITY\;(Case\;b) = \frac{18}{27}\\\\PROBABILITY\;(Case\;b) = 0.666$

c) All Members Assigned to a Different Station.

For this scenario, we have the following results:

[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).

Therefore,

$PROBABILITY\;(Case\;c) = \frac{6}{27}\\\\PROBABILITY\;(Case\;c) = 0.222$

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2 years ago

A food truck sells sandwiches and

anyanavicka [17]

the answer: is $28 dollars

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1 year ago

Name two ways you might see decimals used outside of school.

slega [8]

In money and weighing things

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2 years ago

Read 2 more answers

Lonzell said the function shown in the graph is positive on the interval (−,) and negative on the interval (−,−)open negative 5

Nezavi [6.7K]

Answer:

The positive graph is on the points (-5,-4) and (2,5)

And, negative would be (-4,2)

Step-by-step explanation:

As it seems that fgraph shows positive in the case when it is over x-axis, and negative when it is below y-axis

As on the interval (-1,5)

The graph is below x-axis i.e. on (-1,2)

And, over x-axis i.e. on (2,5)

At this point Lonzell’s is not correct

As on the interval (-5,-1)

The graph is below x-axis i.e. on (-4,-1)

And, over x-axis i.e. on (-5,4)

At this point Lonzell’s is not correct

So,

The positive graph is on the points (-5,-4) and (2,5)

And, negative would be (-4,2)

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2 years ago

Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th

Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

(a)

The bid should be greater than $10,000 to get accepted by the seller. Let bid x be a continuous random variable that is uniformly distributed between

$10,000 and $15,000

The interval of the accepted bidding is $[ {\rm{\$ 10,000 , \$ 15,000}]$ , where b = $15000 and a = $10000.

The interval of the provided bidding is [$10,000,$12,000]. The probability is calculated as,

$\begin{array}{c}\\P\left( {X{\rm{ < 12,000}}} \right){\rm{ = }}1 - P\left( {X > 12000} \right)\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{12000}^{15000}\\\end{array}$

$=1- \frac{[15000-12000]}{5000}\\\\=1-0.6\\\\=0.4$

(b) The interval of the accepted bidding is [$10,000,$15,000], where b = $15,000 and a =$10,000. The interval of the given bidding is [$10,000,$14,000].

$\begin{array}{c}\\P\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{14000}^{15000}\\\end{array} P(X14000)$

$=1- \frac{[15000-14000]}{5000}\\\\=1-0.2\\\\=0.8$

(c)

The amount that the customer bid to maximize the probability that the customer is getting the property is calculated as,

The interval of the accepted bidding is [$10,000,$15,000],

where b = $15,000 and a = $10,000. The interval of the given bidding is [$10,000,$15,000].

$\begin{array}{c}\\f\left( {X = {\rm{15,000}}} \right){\rm{ = }}\frac{{{\rm{15000}} - {\rm{10000}}}}{{{\rm{15000}} - {\rm{10000}}}}\\\\{\rm{ = }}\frac{{{\rm{5000}}}}{{{\rm{5000}}}}\\\\{\rm{ = 1}}\\\end{array}$

(d) The amount that the customer bid to maximize the probability that the customer is getting the property is $15,000, set by the seller. Another customer is willing to buy the property at $16,000.The bidding less than $16,000 getting considered as the minimum amount to get the property is $10,000.

The bidding amount less than $16,000 considered by the customers as the minimum amount to get the property is $10,000, and greater than $16,000 will depend on how useful the property is for the customer.

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2 years ago