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aleksley [76]
3 years ago
15

The school soccer coach records the number of goals some players have scored so far this season. The dot plot of the number of g

oals is shown below.
What is the mean absolute deviation of the data set?


Identify all of the striking deviations.

Mathematics
1 answer:
yuradex [85]3 years ago
3 0

Answer:

the means absolute deviation is 1.8

the striking deviation is 7

(it is because the data point on the far right of the graph)

Step-by-step explanation:

calculate the absolute deviation:

1.  calculate the mean (add all numbers and divide by 10) = 2.6

2.  find the absolute value of each

2.6 - 0 = 2.6

2.6 - 1 = 1.6 (you have 1's 4 times so each one will be 1.6)

2.6 - 3 = 0.4

2.6 - 4 = 1.4 (you have 4's 3 times)

2.6 - 7 = 4.4

3.  Then add all of those values together

2.6 + 1.6 + 1.6 + 1.6 +  1.6 + 0.4 + 1.4 + 1.4 + 1.4 + 7.4 = 18

4.  find the mean of the difference

18/10 = 1.8

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Divide 890 by 1,000,000. (Do not round off)
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Answer:

0.00089‬

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
First, Diana scores 12 points in total with three arrows. On her second turn, she scores 15 points. How many points does she sco
kotegsom [21]

Answer:

Third turn point = 21 points

Step-by-step explanation:

Given:

First turn point = 12

Second turn point = 15

Find;

Third turn point

Computation:

First turn point = 12

Assume;

Point for outer circle = a

Point for inner circle = b

So,

3a = 12

Point for outer circle = 4 point

Second turn point = 15

So,

2a + b = 15

2(4) + b = 15

Point for inner circle = 7

Third turn point = 3 x 7

Third turn point = 21 points

6 0
3 years ago
Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m
Elenna [48]

Answer:

Part a: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b: <em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c: <em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d: <em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

Substitute the value of λ=10 in the formula as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}

​Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b:

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}

<em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as \frac{X}{4}

\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}

That is,

Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)

So, the probability mass function of Y is,

P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d:  

The probability that there is at least one arrival in a 15-second period is calculated as,

\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}

            \begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}

<em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

​

​

7 0
3 years ago
2 less than four times a number is equal to twice a number increase by 8 write an equation and solve to find the value of x
Tom [10]
4x - 2 = 2x + 8
2x -2 = 8
2x = 10
x = 5

4(5)-2 = 2(5)+8
20-2 = 10+8
18 = 18
5 0
3 years ago
Show the reflection of the given points.Locate the points of the reflection in their proper position around the x-axis
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Answer: I believe this is it

4 0
2 years ago
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