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Katyanochek1 [597]
2 years ago
5

Help plz I will give brainliest!!!

Mathematics
1 answer:
Eduardwww [97]2 years ago
3 0
I think c would be the correct answer
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Question 9 (1 point)
Romashka [77]

Answer:

Step-by-step explanation:

League A                   League B

151.12                            163.25

148                                157

26.83                             24.93

29                                  136

136                                145

167                                178

207                               256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{26.83+29 +136+ 148+ 151.12+ 167+207}{7}\\\\Mean =123.564

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}==\sqrt{\frac{(26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2}{7}}=63.98

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

(82.5-1.5\times 76.56,159.06+1.5\times 76.56)

(-32.34,273.9)

So, There is no outlier

Maximum = 207

2)

League B in ascending order :

24.93,136,145,157,163.25,178,256

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{24.93+136+145+157+163.25+178+256}{7}\\\\Mean =151.45

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}==\sqrt{\frac{(24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2}{7}}=68.42

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 24.93,136,145,157

n = 4

Median = \frac{\frac{n}{2} \text{th term}+(\frac{n}{2}+1) \text{th term}}{2}\\Median = \frac{\frac{4}{2} \text{th term}+(\frac{4}{2}+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = \frac{136+145}{2}=140.5

Q1=140.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 157,163.25,178,256

n = 4

Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

(140.5-1.5\times 30.125,170.625+1.5\times 30.125)

(95.3125,215.8125)

24.93 and 256 are outliers  

Maximum = 256

5 0
3 years ago
You roll a 6-sided die.<br> What is P(even or factor of 45)?<br> Write your answer as a percentage.
jonny [76]

Answer:

30% because the chance of the whole dice is 6

4 0
2 years ago
Read 2 more answers
The sum of two positive integers, a and b, is at least 30. The difference of the two integers is at least 10. If b is th
Dafna11 [192]

Answer:

a + b \geq 30

b - a \geq 10

Step-by-step explanation:

Given

The sum of the two positive integer a and b is at least 30, this means the sum of the two positive integer is 30 or greater than 30, so we write the inequalities as below.

a + b \geq 30

The difference of the two integers is at least 10, if b is the greater integer then we subtract integer a from integer b, so we write the inequality as below.

b - a \geq 10

Therefore, the following system of inequalities could represent the values of two positive integers a and b.

a + b \geq 30

b - a \geq 10

6 0
3 years ago
Look at the picture. I really need this answer to continue and I just can’t figure it out! I must be doing something wrong
kompoz [17]
Volume of a cone = pir^2h/3
so substitute what you know

1,102.14 = 3.14(r^2)(13/3)
divide by 3.14
351 = r^2(4.3333)
divde by 13/3
81 = r^2
r = 9 inches


6 0
3 years ago
Ok y’all i need to know something.
Vladimir [108]

Answer:

yah if all your grades are A's and B's

Step-by-step explanation:

6 0
3 years ago
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