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4 years ago

using a fair coin and a fair six-sided number cube, what is the probability of tossing tails and rolling a multiple of 3?

2 answers:

8 0

The multiple of 3 in a fair dice is 3 and 6, thus the probability of obtaining a multiple of 3 will be:
P(3 or 6)=1/6+1/6=1/3

Given a fair coin:
P(Tails)=1/2
thus
<span>probability of tossing tails and rolling a multiple of 3
=1/2</span>×1/3=1/6

stiv31 [10]4 years ago

3 0

$|\Omega|=2\cdot6=12\\ |A|=1\cdot2=2\\\\ P(A)=\dfrac{2}{12}=\dfrac{1}{6}\approx17%$

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Can I get some help? I will be giving brainliest!

vfiekz [6]

Answer:

1.) slope = 2

2.) slope = $-2\frac{2}{3}$

3.) slope = $\frac{2}{3}$

4.) slope = $\frac{1}{4}$

5.) slope = $1\frac{8}{9}$

6.) slope = -2

7.) slope = -3

Step-by-step explanation:

In order to find the slope we have to divide they "y" value (the number of units by which the piece of antler goes up) by the "x" value the number of units by which the piece of antler goes to the side). Notice the if the antler moves to the left as it foes up then the "x" value is negative.

Now as I said before we know this......

Let "m" be the slope, then....

$m = \frac{y}{x}$

So now we just have to plug in the y and x values for each antler and we will get the slopes for each one of them.

1.)

$m = \frac{y}{x}$

$m= \frac{4}{2}$

$m = 2$

2.)

$m = \frac{y}{x}$

$m = \frac{4}{-1.5}$

$m = -2\frac{2}{3}$

3.)

$m = \frac{y}{x}$

$m = \frac{2}{3}$

4.)

$m = \frac{y}{x}$

$m = \frac{1}{4}$

5.)

$m = \frac{y}{x}$

$m = \frac{8.5}{4.5}$

$m = 1\frac{8}{9}$

6.)

$m = \frac{y}{x}$

$m = \frac{2}{-1}$

$m= -2$

7.)

$m = \frac{y}{x}$

$m = \frac{3}{-1}$

$m = -3$

3 0

3 years ago

The fraction 1/25 as a decimal

umka2103 [35]

Answer:

0.04

Step-by-step explanation:

there you go now can you help me???????...

8 0

3 years ago

Read 2 more answers

What are the y and x intercepts of y=3(x-6)²-9 and y=-2(x+5)²+3 .. Thanks

JulsSmile [24]

1.
$y=3(x-6)^2-9 \\ \\
\hbox{x-intercept - y=0} \\
0=3(x-6)^2-9 \\
9=3(x-6)^2 \ \ \ |\div 3 \\
3=(x-6)^2 \\
\sqrt{3}=\sqrt{(x-6)^2} \\
\pm \sqrt{3}=x-6 \\
\pm \sqrt{3}+6=x \\
x=6+\sqrt{3} \ \lor \ x=6-\sqrt{3}} \\ \\
\hbox{y-intercept - x=0} \\
y=3(0-6)^2-9 \\
y=3 \times (-6)^2-9 \\
y=3 \times 36-9 \\
y=108-9 \\
y=99 \\ \\
\boxed{\hbox{x-intercepts: } 6+\sqrt{3} \hbox{ and } 6-\sqrt{3}} \\
\boxed{\hbox{y-intercept: } 99}$

2.
$y=-2(x+5)^2+3 \\ \\
\hbox{x-intercept - y=0} \\
0=-2(x+5)^2+3 \\
-3=-2(x+5)^2 \ \ \ |\div (-2) \\
\frac{3}{2}=(x+5)^2 \\
\sqrt{\frac{3}{2}}=\sqrt{(x+5)^2} \\
\pm \sqrt{\frac{3}{2}}=x+5 \\
\pm \sqrt{\frac{3}{2}}-5=x \\
x=\sqrt{\frac{3}{2}}-5 \ \lor \ x=-\sqrt{\frac{3}{2}}-5 \\ \\
\hbox{y-intercept - x=0} \\
y=-2(0+5)^2+3 \\
y=-2 \times 5^2 +3 \\
y=-2 \times 25+3 \\
y=-50+3 \\
y=-47 \\ \\ \boxed{\hbox{x-intercepts: } \sqrt{\frac{3}{2}}-5 \hbox{ and } -\sqrt{\frac{3}{2}}-5} \\ \boxed{\hbox{y-intercept: } -47}$

7 0

3 years ago

The ratio of height of a bonsai ficus tree to the height of a full-size ficus trees is 1:9 the bonsai ficus is 6 inches tall wha

kotykmax [81]

To solve this relationship simply make them equal to each other and find the correct value to multiply both the top and cotton numbers and solve,
In this case because of the initial ratio of 1:9, if the bonsai ficus is 6 then 54 inches would be the full size.

8 0

3 years ago

Graph f(x) = x2 + 2x - 3, label the function’s x-intercepts, y-intercept and vertex with their coordinates. Also draw in and lab

Semenov [28]

<h2>Answer with explanation:</h2>

The given function : $f(x)=x^2+2x-3$