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Mazyrski [523]
3 years ago
12

Izaiah is taking a drivers

Mathematics
2 answers:
True [87]3 years ago
6 0
64 questions are on the test
yKpoI14uk [10]3 years ago
4 0

64 or 48, im pretty sure it is 64 because it doesn't say " how many are left" it says "how many questions" as in how many in total

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7x/2+5=8<br><br> solve for x
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X= 6/7

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5a + 7b - 12a + 6c - 13b - 22a + 53c
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-29a-6b+59c

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An elevator goes up 7 floors and then down 4 floors. What integer represents the change in the floor level?
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7-4=3
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Heather has a rectangular yard, she measures it and finds out it is 24 1/2 feet long by 12 4/5 feet wide. She wants to know how
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Answer: $2,446.08

Step-by-step explanation:

Lets convert the length and width into decimal form so it's easier to understand.

24 1/2= 24.5

12 4/5= 12.8

Now we must multiply to get the area. 24.5x12. 8= 313. 6 ft

Each square inch costs .65 so we must multiply .65 by 12 to see how much a foot costs. 12x0.65= 7.8

Now that we know how much each foot cost lets multiply 313.6 by 7.8. 313.6x 7.8= $2446.08

3 0
2 years ago
a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade
12345 [234]

Answer:

P(at least 1 dull blade)=0.7068

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.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

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

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

6 0
2 years ago
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