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3 years ago

What's the perimeter of a rectangle with length 12 m and width 5 m? A. 49 m B. 34 m C. 17 m D. 60 m

2 answers:

7 0

Since a rectangle has two sets of equal pairs- two sets of 12m lines and two sets of of 5m lines.
So you add the two 12m lines to get 24m, and then the two 5m lines to get 10m.
then you add 24 and 10, giving you 34m as a perimeter of 34m or answer B!

Leya [2.2K]3 years ago

6 0

B.34m u add 12+12+5+5=34m or u cadn do 12x2+5x2=34m

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Yuki888 [10]

Answer:

Step-by-step explanation:

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4 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

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3 years ago

Melissa reserves another rectangular patch in her vegetable garden for growing bell peppers. Since the space that she can reserv

Otrada [13]

Answer:

Quadratic, 16, 18, 6

Step-by-step explanation:

Did this assignment myself

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3 years ago

A single card is drawn from a 52 standard deck. Find the conditional probability that the card is a heart given that it's a jack

MakcuM [25]

You can use the formula for conditional probability as Puzzling did. However in this problem it is easier just to restrict yourself and enumerate the cases that are left.

There are 4 jacks. One of them is a heart. So the probability is 1/4. 

If you want to look at what the formula for conditional probability does, it adjusts your probability by restricting the "universe". 

P(heart and jack)/P(jack) 
= (1/52)/(4/52) = 1/4

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3 years ago

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balu736 [363]

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