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koban [17]
3 years ago
12

S(t)=16t^2+32t, how long will it take the ball to fall 79 feet?

Mathematics
1 answer:
Phantasy [73]3 years ago
5 0
Mudando o sinal:
-16t² - 32t = 0
-16t*(t-2) = 0
16t = 0 --> t = 0
t-2 = 0 --> t = 2
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What expression is equivalent to 18c^8 d^9/9c^3 d^6
Annette [7]
First, since all of the terms are multiplied, you can cancel them out to simplify them:

18/9 = 2

When a term has exponents on the top and bottom of a fraction, you subtract the bottom from the top:

c^8/c^3 = c^5

d^9/d^6 = d^3

So now, we can get rid of the fraction bar and put all of these terms into one expression:

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7 0
4 years ago
A production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A
viva [34]

Answer:

a)  Select ( 5 Day workers ) = 252,  P ( 5 Day Workers ) = 0.006

b) P ( 5 same shift )  = 0.0074612

c)  P ( At-least 2 different shifts ) = 0.9925

d) P ( Only 2 different shifts )  = 0.3366

Step-by-step explanation:

Given:-

- The number of day shift workers D = 10

- The number of swing shift workers S = 8

- The number of graveyard shift workers G = 6

- The total selection made by the Quality Team = 5

Find:-

(a) How many selections result in all 5 workers coming from the day shift?What is the probability that all 5 selected workers will be from the day shift?

Solution:-

- To select the slips such that all 5 are for Day shift workers, in other words you are also selecting 5 Day shift workers from a pool of 10 Day shift workers. The following number of combinations would be:

                          Select ( 5 Day workers ) = 10 C 5 = 252 combinations

- The total possible outcomes for selecting 5 workers from any of the shifts is:

                         Select ( 5 Workers ) = 24 C 5 = 42504 combinations

- The associated probability for selecting 5 day shift workers is:

          P ( 5 Day Workers ) = Select ( 5 Day workers ) /  Select ( 5 Workers )

                                             = 252 / 42504

                                             = 0.006

Find:-

(b) What is the probability that all 5 selected workers will be from the same shift? (Round your answer to four decimal places.)

Solution:-

- To select the slips such that all 5 are from same shift, in other words you are selecting 5 Day shift workers, or 5 Swing shift workers or 5 Graveyard shift workers from a pool of 10 Day shift workers, 8 Swing shift workers, 6 graveyard shift workers. The following number of combinations would be:

                          Select ( 5 Day workers ) = 10 C 5 = 252 combinations

                          Select ( 5 Swing workers ) = 8 C 5 = 56 combinations

                          Select ( 5 Graveyard workers ) = 6 C 5 = 6 combinations

- The associated probability for selecting 5 same shift workers is:

  P ( 5 Day Workers ) = Select ( 5 Day workers ) /  Select ( 5 Workers )

                                             = 252 / 42504

                                             = 0.006

 P ( 5 Swing Workers ) = Select ( 5 swing workers ) /  Select ( 5 Workers )

                                             = 56 / 42504

                                             = 0.00132

P ( 5 Graveyard Workers ) = Select ( 5 graveyard workers ) /  Select ( 5 Workers )

                                             = 6 / 42504

                                             = 0.0001412

- P ( 5 same shift )  = P ( 5 Day Workers ) + P ( 5 Swing Workers ) + P ( 5 Graveyard Workers )

                                = 0.006 + 0.00132 + 0.0001412

                                =  0.0074612

Find:-

(c) What is the probability that at least two different shifts will be represented among the selected workers?

Solution:-

- To select the slips such that all 5 are from different shifts, in other words you are selecting either a combination of Day shift workers and Swing shift workers or Day shift and Graveyard shift workers or Swing shift and Graveyard shift workers or a combination of all 3. It would be easier if we subtract the probability of no different workers from 1 to get at-least 2 different workers probability. As follows:

- P ( At-least 2 different shifts )  = 1 - P ( 5 same shift Workers )

                                                      = 1 - 0.0074612

                                                      = 0.9925

Find:-

(d) What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

Solution:-

- To select the slips such that all 5 are from only 2 different shifts, in other words you are selecting either a combination of Day shift workers and Swing shift workers or Day shift and Graveyard shift workers or Swing shift and Graveyard shift workers but not a combination of all 3. It would be easier if we subtract the probability of all different workers from the probability of at-least 2 different workers. As follows:

- P ( Only 2 different shifts )  = P ( At-least 2 different ) - P ( All 3 shift Workers )

- The possible combinations for all 3 different shift workers is:

Select ( 3 D ,  1 S , 1 G ) = 10 C 3 * 8 * 6 = 5,760

Select ( 2 D ,  2 S , 1 G ) = 10 C 2 * 8 C 2 * 6 = 7,560

Select ( 2 D ,  1 S , 2 G ) = 10 C 2 * 8  * 6 C 2 = 5,400

Select ( 1 D ,  2 S , 2 G ) = 10 * 8 C 2 * 6C2 = 4,200

Select ( 1 D ,  3 S , 1 G ) = 10 * 8 C 3 * 6 = 3,360

Select ( 1 D ,  1 S , 3 G ) = 10 * 8  * 6 C 3 = 1,600

Total All 3 different shifts selected  = 27,880

P ( All 3 shift Workers ) = 27,880 / 42504 = 0.655938264

- Hence,

P ( Only 2 different shifts ) = 0.9925 - 0.655938264

                                                = 0.3366

6 0
3 years ago
Describe a real world problem that uses the expression 29 divided by (3/8 + 5/6).
Westkost [7]
<span>29 divided by (3/8 + 5/6) 

24</span>
5 0
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The following list has a median of 43. what would be the new median if 21 was added to the list?
Debora [2.8K]

Answer:

38

Step-by-step explanation:

cross the numbers out

8 0
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