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4 years ago

What shape has 4 sides that are different lengths

2 answers:

7 0

Theres no real name for a shape with uneven sides. I just realized that. A four sided figure is a quadrilateral. Or irregular polygon/quadrilateral. A four sided shape with no even sides.

Naya [18.7K]4 years ago

4 0

There is none it would be none

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12.5 feet long weighs 158 pounds. How much does a piece of the same steel pipe 18.75

AveGali [126]

Answer:

237 pounds

Step-by-step explanation:

237 pounds

4 0

3 years ago

HELP PLEASE!!!!!! Anaya bought a sweater that was on sale. The sweater originally cost $34.50 but was purchased for $27.60. What

Elan Coil [88]

Answer: 80%. Hope this helps, please consider making me Brainliest.

Step-by-step explanation:

To find the percentage, divide the sale cost by the original cost:

27.60/34.50

Let's multiply both sides by 10 to make the operation easier:

27.6/34.5 ( I eliminated the zero's because they kind of have no use) -->

276/345, now solve:

276/345 = 0.8

0.8 = 80%

The percentage is 80%.

5 0

1 year ago

Can someone can help me plssssssss

mario62 [17]

Answer:

Rectangle

Step-by-step explanation:

Because the floor is made of floor

5 0

3 years ago

How do you find the slope of a line from a graph that does not go through the origin?

Thepotemich [5.8K]

You find it the same way just with those new coordinates

7 0

3 years ago

A computer programmer has an 18% chance of finding a bug in any given program. What is the probability that she does not find a

LiRa [457]

Answer: $(\frac{41}{50})^{10}$

Step-by-step explanation:

Since, According to the question,

The chance of finding a bug in a program = 18%

Thus, the probability of finding the bug in first attempt = $\frac{18}{100}=\frac{9}{50}$

⇒ The probability of not finding any bug in first attempt = $1 - \frac{9}{50} =\frac{41}{50}$

Similarly, in second attempt , third attempt, fourth attempt_ _ _ _ _ tenth attempt, the probability of not finding any bug is also equal to $\frac{41}{50}$

Thus, the probability that she does not find a bug within the first 10 programs she examines

= $\frac{41}{50}\times \frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}$

= $(\frac{41}{50})^{10}$

5 0

3 years ago