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HACTEHA [7]
3 years ago
7

The probability of getting a 6 after rolling a fair dice 16.

Mathematics
1 answer:
Butoxors [25]3 years ago
4 0

Answer:

P(6) = 5

Step-by-step explanation:

The dice is cast 30 times (k = 30)

So if the big numbers apply, then you would expect the outcome to be evenly distributed, meaning

5 x 1

5 x 2

5 x 3

5 x 4

5 x 5

5 x 6

____ +

30 casts in total (evenly distributed over all numbers of the dice)

P(6) = 1/6 * 30

P(6) = 5

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Luna entered a raffle for a new pair of running shoes. She purchased 20 tickets and was told that she had a 20%
ValentinkaMS [17]

Answer:

She had to sell 4 tickets

Step-by-step explanation:

Do 20 times 20%

20% OF 20

5 0
4 years ago
Read 2 more answers
1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is
sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by d(t)=\frac{t^{2}}{6}+4t where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}

so let's take the first interval:

(a) [1.95, 2.05]

V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}

we get that:

d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125

d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167

so:

V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s

(b) [1.995, 2.005]

V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}

we get that:

d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831

d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835

so:

V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s

(c) [1.9995, 2.0005]

V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}

we get that:

d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358

d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358

so:

V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s

(d) [2, 2.00001]

V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}

we get that:

d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333

d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333

so:

V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

8 0
3 years ago
Simplify 3/4-1/3+1/12​
Semenov [28]

Answer: 1/2 I believe :-)

5 0
3 years ago
Read 2 more answers
Can someon help mee!!
zvonat [6]

I believe it’s A

Sorry if I get it wrong :/

3 0
3 years ago
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Write the first 6 terms of the sequence whose n^th term is 3^n
Ann [662]

Answer:

3, 9, 27, 81, 243, 729

Step-by-step explanation:

3^1 = 3

3² = 9

3³ = 27

3^4 = 81

3^5= 243

3^6= 729

7 0
3 years ago
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