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Ket [755]
2 years ago
7

In college basketball, some teams get most of their points from just one player, and other teams are more balanced in scoring. C

onsider the following facts about the scoring distributions for some NCAA Division I basketball teams. Each team has 121212 players. At Baylor University, there are 666 players who each score 121212 points per game, and there are 666 players who each score 000 points per game. At the University of Maryland, 111 player scores 585858 points per game, 111 player scores 141414 points per game, and the rest of the players score 000 points per game. At Dartmouth College, 444 players score 555 points per game, 444 players score 666 points per game, and 444 players score 777 points per game.
Which of the following is true? Choose all answers that apply: Choose all answers that apply:

(Choice A) A At Dartmouth, the mean number of points per player per game is greater than the median number of points per player per game.
(Choice B) B At Maryland, the mean number of points per player per game is greater than the median number of points per player per game.
(Choice C) C The median number of points per player per game is the same at all 333 schools.
(Choice D) D The mean number of points per player per game is the same at all 333 schools.
Mathematics
1 answer:
lutik1710 [3]2 years ago
7 0

Answer:

(B) At Maryland, the mean number of points per player per game is greater than the median number of points per player per game.

(D)The mean number of points per player per game is the same at all 3 schools.

Step-by-step explanation:

<u>Baylor University</u>

6 players who each score 12 points per game

6 players who each score 0 points per game.

The Scores are: 0,0,0,0,0,0,12,12,12,12,12,12

Mean=\frac{0+0+0+0+0+0+12+12+12+12+12+12}{12} \\Mean=6\\Median=\frac{12+0}{2}=6

<u>University of Maryland</u>

1 player scores 58 points per game,

1 player scores 14 points per game,

The rest(10) of the players score 0 points per game.

The Scores are: 0,0,0,0,0,0,0,0,0,0,14,58

Mean=\frac{0+0+0+0+0+0+0+0+0+0+14+58}{12} \\Mean=6\\Median=\frac{0+0}{2}=0

<u> Dartmouth College</u>

4 players score 5 points per game,

4 players score 6 points per game,

4 players score 7 points per game.

The Scores are: 5,5,5,5,6,6,6,6,7,7,7,7

Mean=\frac{5+5+5+5+6+6+6+6+7+7+7+7}{12} \\Mean=6\\Median=\frac{6+6}{2}=6

The following applies:

(B) At Maryland, the mean number of points per player per game is greater than the median number of points per player per game.

(D)The mean number of points per player per game is the same at all 3 schools.

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For three minutes the temperature of a feverish person has had negative first derivative and positive second derivative.
nydimaria [60]

Answer:

A.) The temperature fell in the last minute, but less than it fell in the minute before.

Step-by-step explanation:

Given that for three minutes the temperature of a feverish person has had negative first derivative and positive second derivative.

i.e. it temperature is represented by T, temperature is variable with

first derivative T' <0 and second derivative T">0

i.e. rate of change of temperature is negative or temperature is falling down in 3 minutes.

But the rate of rate of change of temperature was positive i.e. the rate of change of temperature is increasing as time increases.

So correct option would be option A

A.) The temperature fell in the last minute, but less than it fell in the minute before.

7 0
2 years ago
According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me
Reil [10]

Answer:

a) \bar X=10.65

Median =\frac{10.7+10.7}{2}=10.7

Mode= 10.7

b) Range = 11.8-8.3=3.5

s= 0.948

c)IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8

Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And if we replace we got: \bar X=10.65

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

Median =\frac{10.7+10.7}{2}=10.7

The mode would be the most repeated value on this case:

Mode= 10.7

Part b

The range is defined as Range =Max-Min and if we replace we got:

Range = 11.8-8.3=3.5

We can calculate the standard deviation with the following formula:

s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And if we replace we got:

s= 0.948

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and Q_1= \frac{10.6+10.7}{2}=10.55

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got Q_3 = \frac{10.9+11.2}{2}=11.05

Then we can find the IQR like this:

IQR = Q_3 -Q_1 = 11.05-10.55=0.5

And we can find the usual limits with:

Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8

Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0
2 years ago
Santos flipped a coin 500 times. The coin landed heads up 225 times. Find the ratio of heads to total number of coin flips. Expr
Whitepunk [10]
Here is how you do it....
You will subtract 500 minus 225 to get the number of tails. The number of tails is 275.
Now, you know your ratio is 225:500. All you need to do is simplify it.
225 divided by 25 is 9 and 500 divided by 25 is 20.
Now, the ratio that you have is 9:20.
6 0
2 years ago
Read 2 more answers
The low temperatures in two cities are being compared. In City 1, the range in temperature is 20°F and the IQR is 7°F. In City 2
Tomtit [17]
I wanna say (C) would be your best bet correct me if I’m wrong
5 0
2 years ago
Read 2 more answers
A rectangular pool 18 meters by 12 meters is surrounded by a walkway of width x
vesna_86 [32]

Answer:

x=3 meters

Step-by-step explanation:

step 1

Find the area of the rectangular pool

A=LW

we have

L=18\ m\\W=12\ m

substitute

A=18(12)=216\ m^2

step 2

Find the area of rectangular pool including the area of the walkway

Let

x ----> the width of the walkway

we have

L=(18+2x)\ m\\W=(12+2x)\ m

substitute

A=(18+2x)(12+2x)

step 3

Find the area of the walkway

To find out the area of the walkway subtract the area of the pool from the area of rectangular pool including the area of the walkway

so

A=(18+2x)(12+2x)-216

step 4

Find the value of x if the area of the walkway equal the area of the pool

so

(18+2x)(12+2x)-216=216

Solve for x

(18+2x)(12+2x)=432\\216+36x+24x+4x^{2}=432\\4x^{2} +60x-216=0

Solve the quadratic equation by graphing

The solution is x=3 meters

see the attached figure

8 0
2 years ago
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