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Ad libitum [116K]
3 years ago
14

Dukes science quiz scores are 99,91,60,94,and 95 . Describe the effect of the outlier on the mean and median

Mathematics
2 answers:
lilavasa [31]3 years ago
3 0
I don't know the effect, but I can tell you the outlier, mean, and the median.
Outlier: 60
Mean: 87.8
Median: 60
Goryan [66]3 years ago
3 0

Solution:

we are given that

Dukes science quiz scores are 99,91,60,94,and 95 .

we have been asked to describe the effect of the outlier on the mean and median.

Mean of the given data=\frac{99+91+60+94+95}{5} \\

Mean of the given data=\frac{439}{5}=87.8 \\

Median of the given data after arranging in the ascending order 60,91,94,95,99 is 94.

The outlier is 60.

The outlier is decreasing the value of Mean by 87.8-60=27.8\\

The outlier is decreasing the value of Median by 94-60=34\\


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8 0
3 years ago
Answer this please it’s for algebra 2
Sergeu [11.5K]

Part a)

P(junior) = (number of juniors)/(number total)

P(junior) = 235/705

P(junior) = 1/3 exactly

P(junior) = 0.33333 approximately

======================================

Part b)

P(freshman and LG) = (number of freshman who have LG)/(number total)

P(freshman and LG) = 70/705

P(freshman and LG) = 14/141 exactly

P(freshman and LG) = 0.09929 approximately

======================================

Part c)

n(junior) = number of juniors = 235

n(samsung) = number of people who have samsung = 274

n(junior and samsung) = number of juniors who have samsung = 93

----

n(junior or samsung) = n(junior)+n(samsung)-n(junior and samsung)

n(junior or samsung) = 235+274-93

n(junior or samsung) = 416

----

P(junior or samsung) = n(junior or samsung)/n(total)

P(junior or samsung) = 416/705 exactly

P(junior or samsung) = 0.59007 approximately

======================================

Part d)

n(sophomore and apple) = number of sophomores who have apple

n(sophomore and apple) = 80

n(apple) = number of students who have apple

n(apple) = 261

P(sophomore | apple) = n(sophomore and apple)/n(apple)

P(sophomore | apple) = 80/261 exactly

P(sophomore | apple) = 0.30651 approximately

======================================

Part e)

define the events

A = junior who has apple

B = junior who has samsung

C = person is a junior

n(A or B) = number of juniors who have apple or samsung

n(A or B) = n(A) + n(B) ... A and B assumed mutually exclusive

n(A or B) = 87+93

n(A or B) = 180

n(C) = 235

P( (Apple or Samsung) | Junior ) = n(A or B)/n(C)

P( (Apple or Samsung) | Junior ) = 180/235

P( (Apple or Samsung) | Junior ) = 36/47 exactly

P( (Apple or Samsung) | Junior ) = 0.76596 approximately

4 0
3 years ago
How many solutions exist for the system of equations graphed below?
tamaranim1 [39]

Considering the given graph, the number of solutions for the system of equations is given by:

b.) one

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In a graph, the number of solutions is given by the number of times the lines intersect. Hence, in this problem, the system has one solution, and option b is correct.

More can be learned about a system of equations at brainly.com/question/24342899

#SPJ1

4 0
2 years ago
Find the equation of the line shown.
Luda [366]

Answer:

y = -x + 9

Step-by-step explanation:

✔️Find the slope using (9, 0) and (7, 2):

slope (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{7 - 9} = \frac{2}{-2} = -1

Slope (m) = -1

✔️Find the y-intercept (b):

The line intercepts the y-axis when y = 9. Therefore, y-intercept (b) = 9

✔️Substitute m = -1 and b = 9 into y = mx + b

Thus,

y = -1(x) + 9

y = -x + 9

5 0
3 years ago
A toy manufacturer wants to know how many new toys children buy each year. A sample of 601 children was taken to study their pur
Lilit [14]

Answer:

The confidence interval is 6.6<μ<6.8.

Step-by-step explanation:

We have:

Number of observations = 601

Mean = 6.7

Standard deviation σ = 1.5

The z-score for a 95% confidence interval is 1.96.

The limits of the confidence interval can be calculated as

X \pm z*\frac{\sigma}{\sqrt{n}}\\\\LL=X-z*\frac{\sigma}{\sqrt{n}}=6.7-1.96*\frac{1.5}{\sqrt{601} } =6.7-0.1199=6.6\\\\UL=X+z*\frac{\sigma}{\sqrt{n}}=6.7+1.96*\frac{1.5}{\sqrt{601} } =6.7+0.1199=6.8

The confidence interval is 6.6<μ<6.8.

7 0
3 years ago
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