MATHEMATICS COLLEGE

A local doctor’s office logged the number of patients seen in one day by the doctor for ten days. Find the means, median, range, and midrange of the patients seem in 10 days. 27 31 27 35 35 25 28 35 33 24

Answers

Answer 1
Answer:

Answer:

Mean = 30, Median = 29.5, Range = 9 and Mid-range = 29.5.

Step-by-step explanation:

We are given that a local doctor’s office logged the number of patients seen in one day by the doctor for ten days.

Arranging the given data in ascending order we get;

24, 25, 27, 27, 28, 31, 33, 35, 35, 35.

(a) Mean is calculated by using the following formula;

         Mean, \bar X  =  \frac{\text{Sum of all values}}{\text{Total number of observations}}

                          =  (27+ 31+ 27+ 35+ 35+ 25+ 28+ 35+ 33+ 24)/(10)

                          =  (300)/(10)  = 30

So, the mean of the given data is 30.

(b) For calculating the median, we have to first have to observe that the number of observations (n) in the data is even or odd.

  • If n is odd, then the formula for calculating median is given by;

                     Median  =  ((n+1)/(2))^(th) \text{ obs.}

  • If n is even, then the formula for calculating median is given by;

                     Median  =  \frac{((n)/(2))^(th) \text{ obs.}+ ((n)/(2)+1)^(th) \text{ obs.} }{2}

Here, the number of observations is even, i.e. n = 10.

So,  Median  =  \frac{((n)/(2))^(th) \text{ obs.}+ ((n)/(2)+1)^(th) \text{ obs.} }{2}

                     =  \frac{((10)/(2))^(th) \text{ obs.}+ ((10)/(2)+1)^(th) \text{ obs.} }{2}

                     =  \frac{(5)^(th) \text{ obs.}+ (6)^(th) \text{ obs.} }{2}

                     =  (28+31)/(2)

                     =  (59)/(2)  =  29.5

So, the median of the data is 29.5.

(c) The range of the data is given by = Highest value - Lowest value

                                                        = 35 - 24 = 9

So, the range of the data is 9.

(d) Mid-range of the data is given by the following formula;

                   Mid-range  =  \frac{\text{Highest value}+\text{Lowest value}}{2}

                                      =  (35+24)/(2) = 29.5
Answer 2
Answer:

Final answer:

The mean of the patients seen in 10 days is 30, the median is 29.5, the range is 11, and the midrange is 29.5.

Explanation:

To find the mean, median, range, and midrange of the numbers, we first need to understand what these terms mean. The mean is the average, the median is the middle number in a sorted set, the range is the difference between the highest and lowest values, and the midrange is the average of the highest and lowest values.

First, let's sort the numbers: 24 25 27 27 28 31 33 35 35 35.

To calculate the mean, add all the numbers and divide by the count (10): (24+25+27+27+28+31+33+35+35+35)/10 = 30.

The median is the average of the two middle numbers (which are 28 and 31 in this case): (28 + 31) / 2 = 29.5.

The range is the highest number minus the lowest number: 35 - 24 = 11.

The midrange is the average of the highest and lowest values: (35 + 24) / 2 = 29.5.

Learn more about Mean, Median, Range

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What happens to the area of a triangle after you translate it 2 units to the left and 1unit up?

Answers

Answer:

nothing

Step-by-step explanation:

If you are translating entire triangle, then translating doesn't influence the area. It's just going to change where the triangle is located not its size

I need help with all four Identify the cords, inscribed angles, and central angles in the figure. The center of the circles in #1,2, and 4 is C.

Answers

Answer:

  see below

Step-by-step explanation:

Any line between two points on the circle is a chord.

Any angle with sides that are chords and with a vertex on the circle is an inscribed angle.

Any angle with sides that are radii and a vertex at the center of the circle is a central angle. Each central angle listed here should be considered a listing of two angles: the angle measured counterclockwise from the first radius and the angle measured clockwise from the first radius.

1.

chords: DE, EF

inscribed angles: DEF

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2.

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inscribed angles: SRT, RSU, RST, RTS, TSU

central angles: RCS, RCT, RCU, SCT, SCU, TCU

3.

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During the period of time that a local university takes phone-in registrations, calls come inat the rate of one every two minutes.
a. What is the expected number of calls in one hour?
b. What is the probability of three calls in five minutes?
c. What is the probability of no calls in a five-minute period?

Answers

Answer:

a) The expected number of calls in one hour is 30.

b) There is a 21.38% probability of three calls in five minutes.

c) There is an 8.2% probability of no calls in a five minute period.

Step-by-step explanation:

In problems that we only have the mean during a time period can be solved by the Poisson probability distribution.

Poisson probability distribution

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) = (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.

a. What is the expected number of calls in one hour?

Calls come in at the rate of one each two minutes. There are 60 minutes in one hour. This means that the expected number of calls in one hour is 30.

b. What is the probability of three calls in five minutes?

Calls come in at the rate of one each two minutes. So in five minutes, 2.5 calls are expected, which means that \mu = 2.5. We want to find P(X = 3).

P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)

P(X = 3) = (e^(-2.5)*(2.5)^(3))/((3)!) = 0.2138

There is a 21.38% probability of three calls in five minutes.

c. What is the probability of no calls in a five-minute period?

This is P(X = 0) with \mu = 2.5.

P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)

P(X = 0) = (e^(-2.5)*(2.5)^(0))/((0)!) = 0.0820

There is an 8.2% probability of no calls in a five minute period.

Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that the highest bid in excess of $10,000 will be accepted. Assume that the competitor's bid x is a is a random variable that is uniformly distributed between $10,000 and $15,000.a. Suppose you bid $12,000. What is the probability that your bid will be accepted? (please show calculations)
b. Suppose you bid $14,000. What is the probability that your bid will be accepted? (please show calculations)
c. What amount should you bid to maximize the probability that you get the property? (please show calculations)d. Suppose you know someone who is willing to pay you $16,000 for the property. Would you consider bidding less than the amount in part (c)? Why or why not?

Answers

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

=1- ([15000-12000])/(5000)\n\n=1-0.6\n\n=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}\nP\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\n\n = 1 - \int\limits_(14000)^(15000) {\frac{1}{{15000 - 10000}}} dx\n\n = 1 - \int\limits_(14000)^(15000) {\frac{1}{{5000}}} dx\n\n = 1 - \frac{1}{{5000}}\left[ x \right]_(14000)^(15000)\n\end{array} P(X<14,000)=1-P(X>14000)

=1- ([15000-14000])/(5000)\n\n=1-0.2\n\n=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}\nf\left( {X = {\rm{15,000}}} \right){\rm{ = }}\frac{{{\rm{15000}} - {\rm{10000}}}}{{{\rm{15000}} - {\rm{10000}}}}\n\n{\rm{ = }}\frac{{{\rm{5000}}}}{{{\rm{5000}}}}\n\n{\rm{ = 1}}\n\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.

1/4÷(-2/3) =3/8 she is right now did she get the answer​

Answers

Answer:

  see below for the working

Step-by-step explanation:

Dividing by a number is the same as multiplying by the inverse of that number.

\displaystyle(\left((1)/(4)\right))/(\left(-(2)/(3)\right))=-(1)/(4)\cdot(3)/(2)=-(3)/(4\cdot 2)=-(3)/(8)

The Point (0,0) is a solution to which of these inequalities?

Answers

Answer:

(0, 0) s the solution of y-4 < 3x-1 as it satisfies the inequality.

Hence, option C is true.

Step-by-step explanation:

Given the point (0, 0)

a)

Putting the point (0, 0) the inequality

y+4 < 3x-1

0+4 < 3(0)-1

4 < -1

This is false as -1 can not be greater than 4

b)

y-1 < 3x-4

Putting the point (0, 0) the inequality

0-1 < 3(0)-4

-1 < -4

This is false as -1 can not be lesser than -4

c)

y-4 < 3x-1

Putting the point (0, 0) the inequality

0-4 < 3(0)-1

-4 < -1

This is true as -4 is lesser than -1

d)

y+4 < 3x+1

Putting the point (0, 0) the inequality

0+4 < 3(0)+1

4 < 1

This is false as 4 can not be lesser than 1

Therefore, (0, 0) s the solution of y-4 < 3x-1 as it satisfies the inequality.

Hence, option C is true.
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