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Trava [24]
3 years ago
7

Your car has 10 gallons of gas. You use 1 gallon every 20 miles you travel. Fill in the table.

Mathematics
2 answers:
Solnce55 [7]3 years ago
5 0

Answer:

(0, 0), (20, 1), (40, 2), (60, 3), (80, 4)

Step-by-step explanation:

Artist 52 [7]3 years ago
3 0
0
1
2
3
4

These are the numbers you are looking for.
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11111nata11111 [884]

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F(x)= 6+ (10/x) what are the inflection points?
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An inflection point is a point on the graph of a function at which the concavity changes. Points of inflection can occur where the second derivative is zero.    

Step-by-step explanation:

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3 years ago
Who is correct Finn or Fiona
suter [353]
Fiona is correct- if a shape dilated by a fraction it’s always smaller, unless in the fraction the bigger number is on top. So if it’s more than 1 the shape gets bigger. And if it’s less than one it always gets smaller. Hope this didn’t confuse you even more.
8 0
2 years ago
Components of a certain type are shipped to a supplier in batches of ten. Suppose that 52% of all such batches contain no defect
koban [17]

Answer:

P ( B0 / D0 ) = 0.59877

P ( B1 / D0 ) = 0.25793

P ( B2 / D0 ) = 0.14329

Step-by-step explanation:

Given:

-  0 be the event that the batch has 0 defectives = (0 ) = 0.52

- 1 be the event that the batch has 1 defectives = (1 ) = 0.28

- 2 be the event that the batch has 2 defectives = (2 ) = 0.2

- Two components are selected

Find:

What are the probabilities associated with 0, 1, and 2 defective components being in the batch under each of the following conditions?

(a) Neither tested component is defective.

Solution:

Let 0 be the event that neither selected component is defective.

- The event 0 can happen in three different ways:

(i) Our batch of 10 is perfect, and we get no defectives in  our sample of two;

                   P(i) = P(B0) = 0.52

(ii) Our batch of 10 has 1 defective, but our sample of two misses them;

                  P ( no defect / B1 ) = P ( no defect ) / P ( B 1 )

                                                  = 9C2 / 10C2 = 0.8

                  P ( ii ) = 0.28*0.8 = 0.224

(iii) Our batch  has 2 defective, but our sample misses them.

                 P ( no defect / B2 ) = P ( no defect ) / P ( B 2 )

                                                  = 8C2 / 10C2 = 56/90

                  P ( iii ) = 0.2*56/90 = 0.124444

- Then,

                 P(Do) = P(i) + P(ii) + P(iii)

                 P(Do) = 0.52 + 0.224 + 0.124444 = 977/1125

We use the general conditional probability formula:

                P ( B0 / D0 ) = P ( B0 & D0 ) / P( D0 )

                P ( B0 / D0 ) = 0.52*1125 / 977 = 0.59877

                P ( B1 / D0 ) = P ( B1 & D0 ) / P( D0 )

                P ( B1 / D0 ) = 0.224*1125 / 977 = 0.25793

                P ( B2 / D0 ) = P ( B2 & D0 ) / P( D0 )

                P ( B2 / D0 ) = 0.12444*1125 / 977 = 0.14329

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The answer is b)$319
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