Shape 2

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

Lucy has some pennies and some nickels. She has a maximum of 15 coins worth a minimum of $0.47combined.If lucky has 6.9999999999

Verdich [7]

Answer:

Step-by-step explanation:

The worth of a penny is 1 cent = 1/100 = $0.01

The worth of a nickel is 5 cents = 5/100 = $0.05

Let x represent the number of pennies that she could have. Lucy has some pennies and some nickels. She has a maximum of 15 coins. If lucky has 6.999999999999999, it means that

6.999999999999999 + x ≥ 15

x ≥ 15 - 6.999999999999999

x ≥ 8

If the 15 coins worth a minimum of $0.47combined, it means that

0.05(6.999999999999999) + 0.01x ≤ 0.47

0.35 + 0.01x ≤ 0.47

0.01x ≤ 0.47 - 0.35

0.01x ≤ 0.12

x ≤ 0.12/0.01

x ≤ 12

Therefore, the possible values for the number of pennies that she could have is 8 ≤ x ≤ 12

6 0

2 years ago

What is the set of x-intercepts of this graphed function ? A.{-3,-1} B. {-3,-1,3} C. {-3,3} D.{-3}

WARRIOR [948]

Answer:

second option

Step-by-step explanation:

The x- intercepts are the values on the x- axis where the graph crosses.

These are

x = - 3, x = - 1 and x = 3

5 0

3 years ago

Help me please on my homework it's due today!

Eva8 [605]

The answer is D. $756

8 0

3 years ago

Read 2 more answers

Help me please!

Ksivusya [100]

Answer:

sum of a triangle adds up to 180 degrees

a+b+c=180

35+52+93=180

6 0

3 years ago