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inn [45]
3 years ago
12

What is the height?? please help

Mathematics
1 answer:
Naddika [18.5K]3 years ago
3 0

Answer:

The height is 12.4 in

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Which line is parallel to the line given below?
vichka [17]

Answer:

B

Step-by-step explanation:

2x+5y=-52...............

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3 years ago
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I need help on this worksheet for my math hw it’s combing like terms?
maw [93]
What you would do is combine the terms with variables, and combine the terms without.

Answers:
2. -2w-22
3. 15z
4. 7-16s
5. 17w
6. 4-6r
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8. 8q
9. -18q-9
10. 12q-3

I hope I got all of these right if not that’s embarrassing haha I did all of them in my head I’m pretty sure they’re right

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3 0
3 years ago
What are the cost of 1 t-shirt and 1 sweatshirt? (separate)
emmainna [20.7K]

X is t shirt is $8

Y is sweatshirt $11

2x+2y=38

3x+1y=35

We will be using the elimination process so you want even them out same value

2x+2y=38

-6x-2y=-70 ( want get rid of the y-value by increasing to other by when both add or subtract it will equal 0)

-4x=-32

Divide both -4

X=8 So 8 dollars a t-shirt so now will fill x with 8

2(8)+2y=38

16+2y=38

-16

2y=22

Divid by 2

Y = 11 so a sweatshirt is 11 dollars

You can check by fill each by the variable

7 0
3 years ago
An urn contains three white balls and two red balls. The balls are drawn from the urn, oneat a time without replacement, until a
Tpy6a [65]

Answer:

p ( X = 1 ) = 0.6 , p ( X = 2 ) = 0.3 , p ( X = 3 ) = 0.1

Verified

E ( X ) = 1.5

Step-by-step explanation:

Solution:-

- An urn contains the following colored balls:

                       Color                  Number of balls

                       White                            3

                       Red                               2

- A ball is drawn from urn without replacement until a white ball is drawn for the first time.

- We will construct cases to determine the distribution of the random-variable X: The number of trials it takes to get the first white ball.

- We have three following case:

1) White ball is drawn on the first attempt ( X =  1 ). The probability of drawing a white ball in the first trial would be:

              p ( X = 1 ) = ( Number of white balls ) / ( Total number of ball )

              p ( X = 1 ) = ( 3 ) / ( 5 )

2) A red ball is drawn on the first draw and a white ball is drawn on the second trial ( X = 2 ). The probability of drawing a red ball first would be:

      p ( Red on first trial ) = ( Number of red balls ) / ( Total number of balls )

      p ( Red on first trial ) = ( 2 ) / ( 5 )

- Then draw a white ball from a total of 4 balls left in the urn ( remember without replacement ).

   p ( White on second trial ) = ( Number of white balls ) / ( number of balls left )

   p ( White on second trial ) = ( 3 ) / ( 4 )

- Then to draw red on first trial and white ball on second trial we can express:

                p ( X = 2 ) =  p ( Red on first trial ) *  p ( White on second trial )

                p ( X = 2 ) =  ( 2 / 5 ) * ( 3 / 4 )

                p ( X = 2 ) =  ( 3 / 10 )  

3) A red ball is drawn on the first draw and second draw and then a white ball is drawn on the third trial ( X = 3 ). The probability of drawing a red ball first would be ( 2 / 5 ). Then we are left with 4 balls in the urn, we again draw a red ball:

   p ( Red on second trial ) = ( Number of red balls ) / ( number of balls left )

   p ( Red on second trial ) = ( 1 ) / ( 4 )    

 

- Then draw a white ball from a total of 3 balls left in the urn ( remember without replacement ).                  

   p ( White on 3rd trial ) = ( Number of white balls ) / ( number of balls left )

   p ( White on 3rd trial ) = ( 3 ) / ( 3 ) = 1

- Then to draw red on first two trials and white ball on third trial we can express:

                p ( X = 3 ) =  p ( Red on 1st trial )*p ( Red on 2nd trial )*p ( White on 3rd trial )

                p ( X = 3 ) =  ( 2 / 5 ) * ( 1 / 4 ) * 1

                p ( X = 3 ) =  ( 1 / 10 )  

- The probability distribution of X is as follows:

    X          1                  2                      3

p ( X )      0.6               0.3                  0.1

- To verify the above the distribution. We will sum all the probabilities for all outcomes ( X = 1 , 2 , 3 ) must be equal to 1.

          ∑ p ( Xi ) = 0.6 + 0.3 + 0.1

                         = 1 ( proven it is indeed a pmf )

- The expected value E ( X ) of the distribution i.e the expected number of trials until we draw a white ball for the first time:

               E ( X ) = ∑ [ p ( Xi ) * Xi  ]

               E ( X ) = ( 1 ) * ( 0.6 ) + ( 2 ) * ( 0.3 ) + ( 3 ) * ( 0.1 )

               E ( X ) = 0.6 + 0.6 + 0.3

               E ( X ) = 1.5 trials until first white ball is drawn.

8 0
4 years ago
Can someone help me? I am stuck on this question I would mean the world if u helped me! and tysm for the people who helped me!
Olin [163]

Answer:

113.69700

Step-by-step explanation:

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