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DENIUS [597]
3 years ago
6

What’s 6.5 inches times 4?

Mathematics
2 answers:
wolverine [178]3 years ago
8 0

Answer:

your answer is 26 inches

AysviL [449]3 years ago
6 0
Hello, the ANWSER to this question is simply 26, explanation the calculator <3 I hope this helps ! Have a nice day
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A piece of string measures 2 16/20. Which of the following is equivalent to the length of the string in meters?
Ivenika [448]

Answer:

H: 2.8

Step-by-step explanation:

first divide 16 by 20 then you get a decimal.

then you add the decimal answer to the whole number (2)

7 0
3 years ago
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HELP PLEASE 80 POINTS
alexgriva [62]

Answer:

  • x = 15/p
  • x = -3

Step-by-step explanation:

<h3>Given</h3>
  • 4(px + 1)=64
<h3>Find</h3>
  • Solve for x
  • Find x when p = -5
<h3>Solution</h3>
  • 4(px + 1) = 64
  • 4(px + 1)/4 = 64/4
  • px + 1 = 16
  • px = 15
  • x = 15/p

<u>When p = -5, substitute p:</u>

  • x = 15/p
  • x = 15/(-5) = -3
4 0
3 years ago
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A pyramid has a base area of 44cm and a height of 12.6 cm what is the volume of the pyramid ?
aliya0001 [1]

Answer:

184.8 cubic cm

Step-by-step explanation:

The formula for the volume of a pyramid is

V = (1/3)(B)(h)      where B is the area of the base, and h is the height

We are given B = 44 cm² andh = 12.6cm  

Plug them into the formula and simplify...(I'm assuming you are allowed to use a calculator for these problems)

V = (1/3)(44cm²)(12.6cm)

<h2>     V = 184.8 cm³  or 184.8 cubic cm</h2><h2 />
7 0
3 years ago
1 4/11 dividedd by 1 1/4
Gwar [14]
The answer is 1 1/11
5 0
3 years ago
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1000 households were surveyed. 275 households own a desktop computer, 455 households own a DVD player, 405 households own two ca
icang [17]

Answer:

Step-by-step explanation:

From the given information,

Suppose

X represents the Desktop computer

Y represents the DVD Player

Z represents the Two Cars

Given that:

n(X)=275

n(Y)=455

n(Z)=405

n(XUY)=145

n(YUZ)=195

n(XUZ)=110

n((XUYUZ))=265

n(X ∩ Y ∩ Z) = 1000-265

n(X ∩ Y ∩ Z) = 735

n(X ∪ Y) = n(X)+n(Y)−n(X ∩ Y)

145 = 275+455 - n(X ∩ Y)

n(X ∩ Y) = 585

n(Y ∪ Z) = n(Y) + n(Z) − n(Y ∩ Z)

195 = 455+405-n(Y ∩ Z)

n(Y ∩ Z) = 665

n(X ∪ Z) = n(X) + n(Z) − n(X ∩ Z)

110 = 275+405-n(X ∩ Z)

n(X ∩ Z) = 570

a. n(X ∪ Y ∪ Z) = n(X) + n(Y) + n(Z) − n(X ∩ Y) − n(Y ∩ Z) − n(X ∩ Z) + n(X ∩ Y ∩ Z)

n(X ∪ Y ∪ Z) = 275+455+405-585-665-570+735

n(X ∪ Y ∪ Z) = 50

c. n(X ∪ Y ∪ C') = n(X ∪ Y)-n(X ∪ Y ∪ Z)

n(X ∪ Y ∪ C') = 145-50

n(X ∪ Y ∪ C') = 95

6 0
3 years ago
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