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3 years ago

N is the set of real numbers that are factors of 12

1 answer:

6 0

We are asked to determine the set of real numbers that contain the factors of 12. 12 is considered as a composite number in which the factors are not only 1 and 12. Factors are considered as numbers in which 12 is divisible by them. In this case, the answer is B. {1, 2, 3, 4, 6, 12}.

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Dave's dog is 1 decade and 2 years 9months 3days old Dave is half as old as his dog how old is dave

victus00 [196]

Answer:

dave is 6 years old depending on which way you are halfling

Step-by-step explanation: the dog is 12 years and 9 months old

dave would be 6 i believe unless its going the other way then i have no clue

a decade is 10 years btw.

3 0

3 years ago

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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

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

How many minutes is it before 12 noon if 68 minutes ago it was three times as many minutes past 10 a.m.?

umka21 [38]

If 68 minutes ago it (then) was three times as many minutes past 10 am (now)

x + 68 = 3 ×
×
(120 - x) →
→
x = 73 min
… if 68 minutes ago it (now) was three times as many minutes past 10 am (then)

x = 3 ×
×
[120 - (x + 68)] →
→
x = 39 min
… if 68 minutes ago it (then) was three times as many minutes past 10 am (then)

x + 68 = 3 ×
×
[120 - (x + 68)] →
→
x = 22 min

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