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3 years ago

3 exampls of discrete variable

Mathematics

2 answers:

mr Goodwill [35]3 years ago

7 0

Number of coin flips

Number of books published

Distance walked, rounded to the nearest kilometer

saw5 [17]3 years ago

3 0

Number of coin flips: 4. (It can't meaningfully be 4.3)
Number of books published: 2. (It can't meaningfully be 2.1)
Distance walked, rounded to the nearest kilometer: 3 (We rounded it, so it can't be 3.2)

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Two students walk in the same direction along a straight path, at a constant speed - one at .90 m/s and the other at 1.90 m/s.a.

anyanavicka [17]

Answer:

a)The second student arrive 7 minutes and 36 seconds sooner

b)They have to walk 598,51 m

Step-by-step explanation:

Speed= Distance/Time

We are looking for time now

Time = Distance/Speed

First student

Time = 780 m/0.90 m/s=866,66 seconds

Second student

Time = 780 m/1.90 m/s=410,52 seconds

Difference=866,66-410,52= 456,14 seconds

In minutes

Difference= 456,14 seconds /60=7 minutes 36 seconds

Distance=Speed x Time

Speed¹ x Time¹ =Speed² x Time²

Time¹= Time² + 5.50 min

5.50 min =350 seconds

Time² =Speed¹ x Time¹ /Speed²

Time² =Speed¹ x (Time² + 350 seconds ) /Speed²

Time² =0.9 x (Time² + 350 seconds ) /1.9

Time² =0,4737 x (Time² + 350 seconds )

Time² =0,4737 Time² + 165,79 seconds

Time² - 0,4737 Time² = 165,79 seconds

0,5263 Time² = 165,79 seconds

Time² = 165,79 seconds/0,5263 =315,01 seconds

Distance=1,9 x 315,01 seconds = 598,51 m

5 0

2 years ago

Two math students were asked to write an exponential growth equation that had a starting value of 300 and a growth rate of 2%. P

Gemiola [76]

Answer:

Pierre is right

Step-by-step explanation:

The correct formula for Exponential growth rate is given as:

y = a( 1 + r) ^t

Where

y = Amount after time t

a = Initial amount

r = Growth rate

t = time

From the question

a = 300

r = 2% = 0.02

Hence, our exponential growth rate =

y = 300( 1 + 0.02)^t

y = 300( 1.02)^t

Therefore, Pierre is right

5 0

2 years ago

Please help with this math problem ASAP!! Will mark brainliest! :)

Novosadov [1.4K]

I think it’s 5 because I think it’s 6

3 0

2 years ago

find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the large

katrin2010 [14]

A generic odd number can be written as

$2k+1,\quad k \in \mathbb{Z}$

Since there is an odd number every two numbers, three consecutive odd numbers will be

$2k+1,\quad 2k+3,\quad 2k+5$

Now let's make up the equations: the sum of the first two is

$(2k+1)+(2k+3)$

And 27 less than 3 times the largest is

$3(2k+5)-27$

These two must be the same, so we have

$(2k+1)+(2k+3)=3(2k+5)-27 \iff 4k+4 = 6k+30-27 \iff 4k+4=6k+3$

Subtracting 4k and 3 from both sides gives

$1=2k \iff k=\dfrac{1}{2}$

Which means that the problem has no solution.

To confirm this hypothesis, we can observe that, on the left hand side, we have the sum of two odd numbers, which is even

On the right hand side, we have an odd number, multiplied by 3 (still odd), take away 27 (still odd).

So, the left hand side is even, and the right hand side is odd. They can't be the same number.

7 0

3 years ago

Pythagorean theorem in 3D.

Mekhanik [1.2K]

Answer:

Step-by-step explanation:

2^2 + 5^2 = s^2

s^2 = 4 + 25

s^2 = 29

s = sqrt(29)

s = 5.4

6 0

2 years ago

3 exampls of discrete variable ​

3 exampls of discrete variable