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mart [117]
2 years ago
6

A parking space shaped like a parallelogram has a base of 17 feet and a height is 9 feet. A car parked in the space is 16 feet l

ong and 6 feet wide. How much of the parking space is not covered by the car?
Mathematics
1 answer:
Art [367]2 years ago
8 0
First, we have to get the area of the parallelogram. A = b x h. 
A = 17 feet x 9 feet
A = 153 square feet.

The area of the car (assume that the car is rectangle) is equal to: A = l x w.
A = 16 x 6
A = 96 square feet.

The space that is not covered by the car is 153 - 96 = 37 square feet.
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The teacher can schedule 3 classes in a night because 4 1/2 divided by 1 1/2 is 3.

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Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

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But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})

P(at least 1 dull blade)=1-0.2931

P(at least 1 dull blade)=0.7068

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3 years ago
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