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zhannawk [14.2K]
2 years ago
7

Here are Barry Bonds's home-run totals, mostly for the San Francisco Giants, during the years 1991 - 2005:25, 34, 46, 37, 33, 42

, 40, 37, 34, 49, 73, 46, 45, 45, 5There are two outliers at opposite ends of this distribution: his 5 home runs in 2005 and his 73 home runs in 2001. Which of the following is a modified box plot for this distribution?

Mathematics
2 answers:
elixir [45]2 years ago
8 0

Answer:

Refer the attached figure.

Step-by-step explanation:

Given : Here are Barry Bonds's home-run totals, mostly for the San Francisco Giants, during the years 1991 - 2005

The data A={25, 34, 46, 37, 33, 42, 40, 37, 34, 49, 73, 46, 45, 45}  

To find : Modified box plot for this distribution?

Solution :

Arranging the data from lowest to highest,

{25,33,34,34,37,37,40,42,45,45,46,46,49,73}  

Find the median of the data  

As the data is even numbers so the median is

M=\frac{40+42}{2}=\frac{82}{2}=41

Median = 41

For Box plot Graph,

q_1=Q(A,0)

q_1=25

q_2=Q(A,0.25)

q_2=34.75

q_3=Q(A,0.5)

q_3=41

q_4=Q(A,0.75)

q_4=45.75

q_5=Q(A,1)

q_5=73

Plotting the Box plot of the data.

Refer the attached figure below.

liraira [26]2 years ago
4 0

Answer:

The box plot of given data is shown below.

Step-by-step explanation:

The given data set is

25, 34, 46, 37, 33, 42, 40, 37, 34, 49, 73, 46, 45, 45, 5

Arrange the data in ascending order.

5, 25, 33, 34, 34, 37, 37, 40, 42, 45, 45, 46, 46, 49, 73

Divide the data in 4 equal parts.

(5, 25, 33), 34,( 34, 37, 37), 40,( 42, 45, 45), 46, (46, 49, 73)

Minimum = 5

Maximum = 73

Q_1=34

Median=40

Q_3=46

The interquartile range of the data set is

IQR=Q_3-Q_1

IQR=46-34=12

Boundaries are:

Minimum=Q_1-1.5(IQR)\Rightarrow 34-1.5(12)=16

Maximum=Q_3-1.5(IQR)\Rightarrow 46+1.5(12)=64

In range of the box plot is 16 to 64. Box starts from 34 and ends at 46 and mark a line inside the box at 40. Plot one dot above 5 and one dot above 73.

Therefore, the box plot of given data is shown below.

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The linear equation to model the company's monthly expenses is y = 2.5x + 3650

<em><u>Solution:</u></em>

Let "x" be the units produced in a month

It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers.

Cost per unit = $ 2.50

The company has monthly operating expenses of $350 for utilities and $3300 for salaries

We have to write the linear equation

The linear equation to model the company's monthly expenses in the form of:

y = mx + b

Cost per unit = $ 2.50

Monthly Expenses = $ 350 for utilities and $ 3300 for salaries

Let "y" be the total monthly expenses per month

Then,

Total expenses = Cost per unit(number of units) + Monthly Expenses

y = 2.50(x) + 350 + 3300\\\\y = 2.5x + 3650

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2 years ago
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Ted is a long distance driver. It took him 30 min longer to drive 275km on the Trans-Canada highway west of Swift Current, Saska
IRISSAK [1]

Speed traveling west = x - 10 km /hr


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A snowboarder leaves an 8-foot-tall ramp with an upward velocity of 28 feet per second. The function h   16t 2  28t 8 gives
mr_godi [17]

The complete question is;

A snowboarder leaves an 8-foot-tall ramp with an upward velocity of 28 feet per second. The function h = -16t² + 28t + 8 gives the height h (in feet) of the snowboarder after t seconds. The snowboarder earns 1 point per foot of the maximum height reached, 5 points per second in the air, and 25 points for a perfect landing. With a perfect landing, how many total points does the snowboarder receive?

Answer:

Total points earned with a perfect landing = 111 points

Step-by-step explanation:

First of all, let's find the maximum height of the given parabolic function h = -16t² + 28t + 8

h_max = c - (b²/4a)

where: a = -16, b = 28, c = 8

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The question says that the snowboarder earns 1 point per foot of the maximum height reached.

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Now, let's find the maximum time in air by solving the equation for h = 0 Thus;

-16t² + 28t +8 = 0

Using quadratic formula,

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Montano1993 [528]

Answer:

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Step-by-step explanation:

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mean, x = 5

standard deviation, σ = 0.9

number of sample, n = 16

now we calculate the probability that the average diameter of those sand dollars is more than 4.73 centimeters.

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                = 0.5910

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