What is the answer to this question?

You roll two fair dice, a green one and a red one.

olga_2 [115]

Answer:

(a) yes

(b) 1/36

(c) 1/36

(d) 1/18

Step-by-step explanation:

(a) yes they are independent as the outcome of one does not affect the outcome of the other.

(b) As the dice are fair, each possible number (1 through 6) has the same probability of being rolled.

P(1 on green die) = 1/6

P(2 on red die) = 1/6

Therefore, P(1 on green die) AND P(2 on red die) = 1/6 × 1/6 = 1/36

(c) Again, as the dice are fair, each possible number (1 through 6) has the same probability of being rolled.

P(2 on green die) = 1/6

P(1 on red die) = 1/6

Therefore, P(2 on green die) AND P(1 on red die) = 1/6 × 1/6 = 1/36

(d) p[(1 on green die and 2 on red die) OR (2 on green die and 1 on red die)

= 1/36 + 1/36

= 2/36

= 1/18

3 0

2 years ago

What's sixty nine thousand one hundred eight written in standred form

Vadim26 [7]

The answer is 69,108

3 0

3 years ago

At a toy factory, toy cars are produced on an assembly line. It takes 2.054 s to attach each of the four wheels. It takes 3.652

IrinaK [193]

Add together the time it takes to put on the four wheels and the hood.
2.054s +3.652s = 5.704s

multiple this answer by 42

5.706s times 42= 239.652s

answer number two is correct

8 0

3 years ago

Ali wrote this first step in how to add 27 12 on a hundred chart

lina2011 [118]

Say the question more clearly please

5 0

3 years ago

Write an expression for the 12th partial sum of the series 3/2+7/3+19/6+... using summation notation

lapo4ka [179]

Answer:

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=73$

Step-by-step explanation:

$First\ term\ of\ the\ series(a_1)=\frac{3}{2}\\\\Second\ term\ of\ the\ series(a_2)=\frac{7}{3}\\\\Third\ term\ of\ the\ series(a_3)=\frac{19}{6}\\\\a_2-a_1=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\\\\a_3-a_2=\frac{19}{6}-\frac{7}{3}=\frac{5}{6}\\\\Hence\ it\ is\ an\ Arithmetic\ Series\ with\ first\ term=\frac{3}{2}\ and\ constant\ difference=\frac{5}{6}$

$a_1=\frac{3}{2}+0\times \frac{5}{6}\\\\a_2=\frac{3}{2}+1\times \frac{5}{6}\\\\a_3=\frac{3}{2}+2\times \frac{5}{6}\\\\.\\.\\.\\a_n=\frac{3}{2}+(n-1)\times \frac{5}{6}\\\\S_n=a_1+a_2+a_3+......+a_n\\\\S_n=(\frac{3}{2}+0\times \frac{5}{6})+(\frac{3}{2}+1\times \frac{5}{6})+(\frac{3}{2}+2\times \frac{5}{6})+....+(\frac{3}{2}+[n-1]\times \frac{5}{6})\\\\S_n=\sum_{i=1}^n [\frac{3}{2}+(i-1)\times \frac{5}{6}]\\\\S_n=(\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+...n\ times)+\frac{5}{6}(1+2+3+4+...+(n-1))\\\\$

$S_n=\frac{3}{2}\times n+\frac{5}{6}\times \frac{n(n-1)}{2}\\\\$

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=\frac{3}{2}\times 12+\frac{5}{6}\times \frac{(12)(12-1)}{2}\\\\S_{12}=18+55\\\\S_{12}=73$

5 0

3 years ago