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natima [27]
3 years ago
14

What are the differences and similarities between the Pythagorean Theorem and the Distance Formula? Which method do you think mi

ght be more useful in solving most problems?
Mathematics
1 answer:
Julli [10]3 years ago
5 0
I personally would use the Pythagorean theorem over distance because it is a lot simpler  
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At 9 AM Monday morning Thomas feels a beaker with water and places it in the corner of the classroom. At 1 PM Tuesday Thomas exa
aev [14]

The level has dropped 11 mms in 22 hours

that is 0.5mm per hour

So the remaining 31 mm will be gone in 31 / 0.5 = 62 hours

62 hours after 11 am Wednesday brings us to   11 am Friday + 14 hours

=  1 am Saturday.   answer

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3 years ago
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What is the value of the expression 2[4(2^3+5)]-4^2
ivolga24 [154]

Answer:

88

Step-by-step explanation:

Solve inside first

2^3=8

8+5= 13

Now multiply by 4

=52

Now by 2

=104

Now this is what we have left

=104-4^2

Make it easy by solving 4^2 which is 16.

104-16

=88

3 0
3 years ago
At noon, the minute and hour hands overlap. In how many hours will they overlap again?
nadya68 [22]

24 hours

Because 24 hours equals 1 day or one rotation of the clock

7 0
3 years ago
Which ordered pair is a solution to the system of inequalities?<br> y&gt;−2 and x+y≤4
Kamila [148]

Answer:-2,-3

Step-by-step explanation:

8 0
2 years ago
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15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0
3 years ago
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