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notsponge [240]
3 years ago
13

An experiment consists of flipping a single coin followed by rolling a six-sided cube (a die).

Mathematics
2 answers:
Over [174]3 years ago
8 0

Answer:\frac{1}{12}

Step-by-step explanation:

Okay, first, you want to find the probability of getting a head, which would be 1/2 because there are two options and we are only looking for one out of the two. Then, the probability of getting a 6 would be 1/6 because there is only one chance of getting a 6 and there are 6 options. Next, you would have to multiply both quantities to find out what the probability would be of getting both. So it would be \frac{1}{2} ×\frac{1}{6}=\frac{1}{12}

Reptile [31]3 years ago
5 0

Probability of getting a head followed by a 6 is 1/12

Step-by-step explanation:

There are 2 possible outcomes of tossing a coin a head and a tail.

The probability is 1 out of 2 each for head and tail.

Probability of getting a 6 when a dice is rolled is 1 out of 6 for each face i.e 1,2,3,4,5,6

So the probability of getting a head followed by 6 is 1/2*1/6 = 1/12

It can also be understood as below.

There can be 12 possible combinations once a coin is tossed and a die is rolled. Head-1, Head-2. Head-3, Head-4, Head-5, Head-6, Tail-1, Tail-2, Tail-3, Tail-4, Tail-5, Tail-6

In the list there is 1 outcome that we need. Head-6. This again gives us the chance of getting Head-6 as 1 out of 12.

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Step-by-step explanation:

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3 years ago
What are the factors of 2x + 3x - 54? Select two options
tiny-mole [99]

Answer:

The answer: The factors are (2x-9) and (x+6).

The Problem:

Factor 2x^2+3x-54

Step-by-step explanation:

So I'm going to do trial factors using the choices to aid me.

Factored form for this problem if it exist will be in the form:

(mx+n)(kx+p).

In general this is what it would look like if we factored any quadratic in terms of x (given the quadratic is not prime but technically you could factor even over the complex numbers).

Let's look at:

(mx+n)(kx+p)

We want to choose k \text{ and } m such that when you multiply them you get 2.  Well those would have to be 2 and 1.

(2x+n)(x+p)

we want to choose n \text{ and } p such that when you multiply them you get -54. Based on the choices we want to get with -9 and 6, or 9 and -6. We don't know the order we want to choose it in either.

For example which of these would work:

(2x-6)(x+9)

(2x+6)(x-9)

(2x-9)(x+6)

(2x+9)(x-6)

We are going to consider only the outer and inner of FOIL since we already know the first times the first is 2x^2 and the last times the last is -54.

Let's test the first one:

(2x-6)(x+9)

Outer:  2x(9)=18x

Inner: -6(x)=-6x

------------------------ADD!

18x-6x=12x

The first choice did not give us the middle term 3x.

Trying the second one would give us the opposite since they are in the same form as previous just the + and - are switched.

Let's look at the third one:

(2x-9)(x+6)

Outer: 2x(6)=12x

Inner: -9(x)=-9x

--------------------------ADD!

12x-9x=3x

This is the winner.

The answer: The factors are (2x-9) and (x+6).

7 0
3 years ago
Read 2 more answers
"10 less 2 times a number"<br> O 10r - 2<br> 0 2r - 10<br> O 2 - 10<br> O 10 - 2r<br> 10 - 2
4vir4ik [10]

the answer to the question is 2r-10

4 0
3 years ago
Read 2 more answers
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