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Alexxx [7]
4 years ago
8

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

Mathematics
2 answers:
Alborosie4 years ago
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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beks73 [17]
The markup is the percent of the original price that it is increased by. To determine the percentage marked up, you will subtract the new price and the original price, and then divide that by the original price.

200-125=$75

75/200=0.375 or 37.5% markup.
5 0
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Question 1 Draw a line through the point (−2,−2) with a slope of 1/4. Draw a line through the point (2, 0) with a slope of 1/4.
Natasha2012 [34]

Answer:

line1:y=\frac{1}{4}x-\frac{3}{2}

line2:  y=\frac{1}{4}x -\frac{1}{2}

Step-by-step explanation:

4 0
3 years ago
Helppp with 2nd and 4th
DiKsa [7]

Answer:

30

Step-by-step explanation:

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3 years ago
Read 2 more answers
Find the zero of each function and state the multiplicity of each zero. Please show all steps.
vodka [1.7K]

Answer:

1. y=(x+3)^3. Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1). Zeros: x=2 multiplicity 2; x=1 multiplicity 1.

3. y=(2x+3)(x-1)^2. Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.


Step-by-step explanation:

1. y=(x+3)^3

y=0\\ (x+3)^3=0\\ \sqrt[3]{(x+3)^3}=\sqrt[3]{0}\\ x+3=0\\ x+3-3=0-3\\ x=-3

Zero: x=-3 multiplicity 3.


2. y=(x-2)^2 (x-1)

y=0\\ (x-2)^2(x-1)=0\\ \left \{ {{(x-2)^2=0} \atop {x-1=0}} \right\\ \left \{ {{\sqrt{(x-2)^2} =\sqrt{0} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x-2=0} \atop {x=1}} \right\\ \left \{ {{x-2+2=0+2} \atop {x=1}} \right\\ \left \{ {{x=2} \atop {x=1}} \right.

Zeros: x=2 multiplicity 2; x=1 multiplicity 1


3. y=(2x+3)(x-1)^2

y=0\\ (2x+3)(x-1)^2=0\\ \left \{ {{2x+3=0} \atop {(x-1)^2=0}} \right\\ \left \{ {{2x+3-3=0-3} \atop {\sqrt{(x-1)^2} =\sqrt{0} }} \right\\ \left \{ {{2x=-3} \atop {x-1=0}} \right\\ \left \{ {{\frac{2x}{2} =\frac{-3}{2} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x=-\frac{3}{2} } \atop {x=1}} \right.

Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

4 0
3 years ago
The vertices of quadrilateral ABCD are A(1,1), B(1,5), C(5,5), and D(7,1). You want to transform ABCD into a parallelogram by on
elixir [45]
The new coordinate of point B would have to be (-1,5). This can be seen when graphing the coordinates you gave, and by substituting 1 for -1 in the coordinates of point B into that same graph
3 0
3 years ago
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