Factoring: 3x^5+6x^3

During a game of pool, Mia hit a ball and it followed the path from B to C and then to D. The angles formed by the path are cong

dybincka [34]

C.43 in because you got to think about it then you haft to add

8 0

3 years ago

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A. A batch of 30 parts contains five defects. If two parts are drawn randomly one at a time without replacement, what is the pro

Zarrin [17]

Answer:

a) For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

For the second trial since the experiment is without replacement we have 29 parts left, and since we select 1 defective from the first trial we have in total 4 defective left, so then the probability of defective for the second trial is : 4/29

And then we can assume independence between the events and we have this probability:

(5/30)*(4/29) =0.16667*0.1379= 0.0230

b) For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

And since we replace the part selected is the same probability for the second trial and then the final probability assuming independence would be:

(5/30)*(5/30) =0.1667* 0.1667= 0.0278

Step-by-step explanation:

For this case we know that we have a batch of 30 parts with 5 defective.

Part a

If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective?

For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

For the second trial since the experiment is without replacement we have 29 parts left, and since we select 1 defective from the first trial we have in total 4 defective left, so then the probability of defective for the second trial is : 4/29

And then we can assume independence between the events and we have this probability:

(5/30)*(4/29) =0.16667*0.1379= 0.0230

Part b

If this experiment is repeated, with replacement, what is the probability that both parts are defective?

For this case on the first trial we have the following probability of selecting a defective part: 5/30. Because we have 5 in total defective at the begin and a total of 30 parts

And since we replace the part selected is the same probability for the second trial and then the final probability assuming independence would be:

(5/30)*(5/30) =0.1667* 0.1667= 0.0278

3 0

3 years ago

2(3 times 3+4 times 4)

Jobisdone [24]

2(3*3+4*4)

2(9+16)

2( 25)

The answer is 50

Hope I helped! (Pick me <span>as brainliest!)</span>

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3 years ago

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Anita is making a curtain to surround a table. She bought 3 1/4 yards of fabric. Her total cost was $13. What was the cost per y

Ratling [72]

3.25y=13. Isolate y by dividing both sides of the equation by 3.25. So 3.25y/3.25=13/3.25. Your answer is therefore 13/3.25, or 4

7 0

3 years ago

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A field is a rectangle with a perimeter of 960 feet. The length is 100 feet more than the width. Find the width and length of th

marta [7]

To start, let x represent the width and x+100 represent the length.

Since the perimeter of a figure is the sum of all the measurements of the side which can be represented by (x+100)+(x+100)+x+x and since you know your perimeter is 960 feet, you can set the expression equal to 960. This would look like this:
(x+100)+(x+100)+x+x=960

Once you have done that, combine any like terms (combine terms with the same variables and raised to the same power together) which would simplify to this:
4x+200=960

Now that you have your like terms simplified, subtract 200 from both sides to get 4x=760 and finally, to solve for x, or find the width, divide both sides by 4 to get x=190.

Now that you have your width, now you must find your length as the question asks to find the dimensions of the rectangular field. To find the length, add 100 to the width of (190) since according to the information given, the length is 100 more than the width. When you add 100 to 190, you should get that your length is 290.

Now that you have your length and width, you can conclude that the dimensions of the field is 190 by 290 feet, which is your answer :)

4 0

2 years ago