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Tpy6a [65]
2 years ago
12

You need 50 necklaces for a craft fair. You begin with 20 necklaces and make 4 necklaces each hour. A. Choose an expression that

represents the number of necklaces you have after h hours.
Mathematics
1 answer:
alina1380 [7]2 years ago
5 0

Answer:

The expression representing Number of the necklaces after h hours is  20+4h.

Step-by-step explanation:

Given:

Number of necklaces in the beginning = 20

Number of necklaces made per hour = 4

Number of necklaces to be made = 50

We need to write the expression that represents the number of necklaces we have after h hours.

Solution:

Let the number of hours be denoted by 'h'.

now we can say that;

Number of the necklaces after h hours can be calculated by Number of necklaces in the beginning plus Number of necklaces made per hour multiplied by number of hours.

framing in equation form we get;

Number of the necklaces after h hours = 20+4h

Hence The expression representing Number of the necklaces after h hours is  20+4h.

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1 year ago
There are three local factories that produce radios. Each radio produced at factory A is defective withprobability .02, each one
diamong [38]

Answer:

The probability is 0.02667

Step-by-step explanation:

Let's call D1 the event that the first radio is defective and D2 the event that the second radio is defective.

So, if we select both radios any factory, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1) = P(D2∩D1)/P(D1)

Taking into account that 0.02 is the probability that a radio produced at factory A is defective, P(D2/D1) for factory A is:

P(D2/D1)_A=\frac{0.02*0.02}{0.02} =0.02

At the same way, if both radios are from factory B, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)_B=\frac{0.01*0.01}{0.01} =0.01

Finally, if both radios are from factory C, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)_C=\frac{0.05*0.05}{0.05} =0.05

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P(A)=1/3

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P(C)=1/3

Then, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)=P(A)P(D2/D1)_A+P(B)P(D2/D1)_B+P(C)P(D2/D1)_C

P(D2/D1) = (1/3)*(0.02) + (1/3)*(0.01) + (1/3)*(0.05)

P(D2/D1) = 0.02667

6 0
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Answer:

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b) A polygon with 21 sides has 432 possible diagonals. False

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n --> number of sides of a polygon.

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d) The measures of the exterior angles of a nonagon, a nine-sided figure, have a sum of 360°. True.

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