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musickatia [10]
3 years ago
7

A machine uses pieces of tape to seal packages. It cuts the tape into pieces exactly 2.375 inches long. If it uses 4 pieces of t

ape for a package, how many packages can it seal with a roll of tape 1,900 inches long?
Mathematics
1 answer:
Paul [167]3 years ago
6 0
If it uses 4 pieces of tape for a package...and each piece is 2.375 inches long, then it uses (4 * 2.375) = 9.5 inches per package.

and if u have 1900 inches of tape, 1900/9.5 = 200 packages can be sealed
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The space allowed for the mascot on a​ school's Web page is 75 pixels wide by 60 pixels high. Its digital image is 500 pixels wi
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Answer:

It will be 60 pixels high and 75 pixels wide

Step-by-step explanation:

\frac{400}{60}=\frac{500}{x}

As

  • The space allowed for the mascot on a​ school's Web page is 75 pixels wide by 60 pixels high
  • Its digital image is 500 pixels wide by 400 pixels high

So, the expression becomes

\frac{400}{60}=\frac{500}{x}

\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c

400x=60\cdot \:500

400x=30000

\mathrm{Divide\:both\:sides\:by\:}400

\frac{400x}{400}=\frac{30000}{400}

\mathrm{Simplify}

x=75

Therefore, it will be 60 pixels high and 75 pixels wide

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Learn more about ratio from brainly.com/question/4287633

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Which situation is best represented by the following equation?
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Answer:

B, i might be wrong but i think its B

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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

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zysi [14]

Answer:

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Step-by-step explanation:

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Danny and Mark and Evans ages combined is 80. What was there combined age three years ago?
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Since there are three kids, 3x3=9. 80-9=71.  Danny and Mark and Evans ages combined is 71 years
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