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wlad13 [49]
3 years ago
12

Helllllpppp I need to find the area

Mathematics
1 answer:
Masja [62]3 years ago
3 0
Divide it into two squares.
The first square is just (1)1, which equals 1 cm^2, and add that to the second square, which is just (2)2, which equals 4, so 1+4=5 cm^2.
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Select all equations that have graphs with the same y-intercept.
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y = 3x - 8

y = (1/3)x - 8

y= 5x - 8

y= 2x - 8

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If a car salesman made $2450 from a 12% commission on a car sale, what is cost of the car he sold?
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I believe it is 20,416.67

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When 250 students were interviewed, 96 of the students thought USC was the best school in the nation. The probability that one o
melomori [17]

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240/96.

Step-by-step explanation:

250-96 =240 so that's 240/96

3 0
3 years ago
The infinite geometric sum formula can be used to write 0.126126126...as a fraction. What is the numerator of this reduced fract
igor_vitrenko [27]

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Step-by-step explanation:


6 0
3 years ago
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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