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Maksim231197 [3]
2 years ago
10

You roll a 6-sided die two times. What is the probability of rolling a 4 and then rolling a 2? (Write your answer as a fraction

or whole number.)
Mathematics
1 answer:
Delicious77 [7]2 years ago
4 0

Answer:

1/36

Step-by-step explanation:

On the first throw, the probability of rolling any particular number, 1 through 6, is 1 out of 6. So the chance of rolling a 4 is 1/6.

On the second roll, your probability of rolling a 2 is 1/6.

The 'trick' is knwoing what to do with those two numbers.

Here's the rule: If events are dependent on one another, you multiply the probabilities.

Any time you see a scenario where X has to happen <u>and then </u>some other thing (X, Y, or whatever) has to happen, the events are dependent.

Probability of rolling a 4 <u>and then</u> rolling a 2 = 1/6 * 1/6 = 1/36

Hope this helps.

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3/5= 3 fifth size strips
You want to find an equivalent fraction with the denominator of ten, correct? Then you would multiply the fraction in a way that will allow you to have a denominator of ten.
So...
3/5*2/2=6/10
You would need 6 tenth size strips.
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Answer for 1 1/2(n-4 1/2)=12​
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Answer:

n=147/22

Step-by-step explanation:

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Jace gathered the data in the table. He found the approximate line of best fit to be y = –0.7x + 2.36.
finlep [7]

Answer:

the answer is:

-0.86

Step-by-step explanation:

3 0
3 years ago
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Mily recycled 66 cans in February, 94 cans in March, 122 cans in April, and 150 cans in May. If this pattern went from January t
Mice21 [21]

Answer:

38 cans

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6 0
3 years ago
Complete the point-slope equation of the line through (1,3) (5,1) y-3=?
MaRussiya [10]

Answer:

\huge\boxed{y-3=-\frac{1}{2}(x-1)}

Step-by-step explanation:

Point-slope is:

y-y_1=m(x-x_1)

m-\text{This represents the slope.}\\\\(x_1,y_1)-\text{This represents the point used in the equation.}

<h2>Our goal: </h2>

We have to complete the point-slope equation of the line through (1,3) (5,1).

--------------------------------------------------------

We have a incomplete equation of the line.

y-3=m(x-x_1)

We need to find the <u>slope</u> of the line, and the <u>value</u> of  x_1.

--------------------------------------------------------

<h3>Finding 'x1':</h3>

It seems that the value of 3 was used to be y_1. This means that the point (1,3) was used for the equation. This means that x_1 would have to be 1.

<h3>Finding Slope:</h3>

Slope is rise over run.

m=\frac{rise}{run}=\frac{y_2-y_1}{x_2-x_1}

We are given the points (1,3) and (5,1).

m=\frac{1-3}{5-1}=\frac{-2}{4}=\frac{-1}{2}=\boxed{-\frac{1}{2}}

The slope is one-half.

--------------------------------------------------------

We now have enough information to complete the point-slope equation.

{\left \{ {{x_1=1} \atop {m=-\frac{1}{2} }} \right.}\\\\y-3=m(x-x_1)\rightarrow\boxed{y-3=-\frac{1}{2}(x-1)}

Our final equation is:

y-3=-\frac{1}{2}(x-1)

6 0
3 years ago
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