A number cube with faces labeled from 1 to 6 will be rolled once.

Mathematics

1 answer:

tresset_1 [31]2 years ago

3 0

Question:

A number cube with faces labeled from 1 to 6 will be rolled once. The number rolled will be recorded as the outcome.

Give the sample space describing all possible outcomes. Then give all of the outcomes for the event of rolling a number greater than 2.

If there is more than one element in the set, separate them with commas.

Answer:

$S = \{1,2,3,4,5,6\}$

$Greater2 = \{3,4,5,6\}$

Step-by-step explanation:

Given

A roll of a 6 sided number cube

Solving (a): The sample space

This implies that we list out all number on the number cube.

So:

$S = \{1,2,3,4,5,6\}$

Solving (b): Outcomes greater than 2

This implies that we list out all number on the number cube greater than 2 i.e. 3 to 6.

So:

$Greater2 = \{3,4,5,6\}$

You might be interested in

20 more than the product of a number n and five

PtichkaEL [24]

Answer:

5n + 20

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

A sample of tritium-3 decayed to 94.5% of its original amount after a year.

Zolol [24]

(a) If $y(t)$ is the mass after $t$ days and $y(0) = A$ then $y(t) = Ae^{kt}.$

$y(1) = Ae^k = 0.945A \implies e^k = 0.945 \implies k = \ln 0.945$

$\text{Then } Ae^{(\ln 0.945)t} = \frac{1}{2}A\ \Leftrightarrow\ \ln e^{(\ln 0.945)t} = \ln \frac{1}{2}\ \Leftrightarrow\ ( \ln 0.945) t = \ln \frac{1}{2} \Leftrightarrow \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 12.25 \text{ years}$

(b)
$Ae^{(\ln 0.945) t} = 0.20 A \ \Leftrightarrow\ (\ln 0.945) t \ln \frac{1}{5}\ \implies \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 28.45\text{ years}$