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Kobotan [32]
3 years ago
15

A bike take 56 minutes to travel 8 km? How many minutes will take to travel 15?​

Mathematics
1 answer:
rjkz [21]3 years ago
3 0

Step-by-step explanation:

Ok, so it takes the bike 7 minutes to go 1 km, so it would take it 105 minutes to go 15

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1. The snow pack on Mt. Hood in the Oregon Cascades was 150 inches on
Damm [24]

Answer:

The rate of change is 10%

Step-by-step explanation:

The computation of the rate of change is as follows:

Given that as on Jan 15, the snow pack is 150 inches

And, the last year, it would decrease by 11 inches a week after jan 15

So, the rate of change is

= 15 ÷ 150

= 10%

8 0
3 years ago
A group teens were surveyed about whether they
algol13

Answer:

B D

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Solve the above que no. 55
aleksandr82 [10.1K]

Answer:

Let \left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right), we proceed to prove the trigonometric expression by trigonometric identity:

1) \left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right) Given

2) \left(1+\frac{\cos^{2}A}{\sin^{2}A} \right)\cdot \left(1+\frac{\sin^{2}A}{\cos^{2}A} \right)   \tan A = \frac{1}{\cot A} = \frac{\sin A}{\cos A}

3) \left(\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A} \right)\cdot \left(\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A} \right)    

4) \left(\frac{1}{\sin^{2}A} \right)\cdot \left(\frac{1}{\cos^{2}A} \right)    \sin^{2}A+\cos^{2}A = 1

5) \frac{1}{\sin^{2}A\cdot \cos^{2}A}

6) \frac{1}{\sin^{2}A\cdot (1-\sin^{2}A)}    \sin^{2}A+\cos^{2}A = 1

7) \frac{1}{\sin^{2}A-\sin^{4}A} Result

Step-by-step explanation:

Let \left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right), we proceed to prove the trigonometric expression by trigonometric identity:

1) \left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right) Given

2) \left(1+\frac{\cos^{2}A}{\sin^{2}A} \right)\cdot \left(1+\frac{\sin^{2}A}{\cos^{2}A} \right)   \tan A = \frac{1}{\cot A} = \frac{\sin A}{\cos A}

3) \left(\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A} \right)\cdot \left(\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A} \right)    

4) \left(\frac{1}{\sin^{2}A} \right)\cdot \left(\frac{1}{\cos^{2}A} \right)    \sin^{2}A+\cos^{2}A = 1

5) \frac{1}{\sin^{2}A\cdot \cos^{2}A}

6) \frac{1}{\sin^{2}A\cdot (1-\sin^{2}A)}    \sin^{2}A+\cos^{2}A = 1

7) \frac{1}{\sin^{2}A-\sin^{4}A} Result

4 0
3 years ago
14 rounded to the nearest tenth
mr_godi [17]
The answer would be 10 because anything under 4 rounds down (including 4 itself, unlike 5, which rounds up. 
5 0
4 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
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