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sergij07 [2.7K]
3 years ago
13

What is the pobability that a number cube with six

Mathematics
1 answer:
Snowcat [4.5K]3 years ago
5 0

Answer: \dfrac{1}{6}

Step-by-step explanation:

We know that the total number of outcomes for fair dice {1,2,3,4,5,6} = 6

Given : Favorable outcome = 5

i.e. Number of favorable outcomes =1

We know that the formula to find the probability for each event is given by :-

\dfrac{\text{Number of favorable outcomes}}{\text{Total outcomes}}

Then, the probability that the six  faces cube will land with the number 5 face up will be :_

\dfrac{1}{6}

Hence, the required probability = \dfrac{1}{6}

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2 years ago
Read 2 more answers
Which would be the best method to use to solve the following equations? Explain your reasoning.
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A. square root property

3x^2-192=0\\x^2-66=0\\x^2=66\\x= -\sqrt{66} or \sqrt{66}

it has one value with x which is x^2 and it cn be easily solved without having to factorise, quadratic formula cant be used as it need ax^2+bx+c=0 format

B. factorising

x^2-x-6=0\\x^2+2x-3x-6=0\\(x+2)(x-3)=0\\x+2=0  and  x-3=0\\x= -2 \\x= 3

i just felt like this was easier to factorise than the other 2 options left

C. Completing the square

x^2-6x-7=0\\(x-3)^2=0\\x= -1\\x= 7

same reason personal preference

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x^2-17x-7=0\\x=\frac{-(-17)+\sqrt{(-17)^2-4(1)(-7)} }{2(1)} \\x=\frac{-(-17)-\sqrt{(-17)^2-4(1)(-7)} }{2(1)}\\x=17.4\\x=-0.402

the 17 kinda threw me off and i didnt wanna get on factorising or doing completing the square so quadratic formal

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2 years ago
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