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3 years ago

You roll a die and then roll the die again. How many possible outcomes are there

Mathematics

2 answers:

siniylev [52]3 years ago

7 0

Answer:

Step-by-step explanation:

If one event has p possible outcomes, and another event has m possible outcomes, then there are a total of p • m possible outcomes for the two events. Rolling two six-sided dice: Each die has 6 equally likely outcomes, so the sample space is 6 • 6 or 36 equally likely outcomes.

nignag [31]3 years ago

4 0

Answer:

There will be 36 possible outcomes and 36 sample spaces

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6x+5y=-15 into intercept form

Yuki888 [10]

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3 years ago

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Give the values of a, b, and c needed to write the equation's general form.. (5 + x)(5 - x) = 7. A = 1; B = 0; C = -18. A = 25;

Licemer1 [7]

The general form of an equation is $A x^{2} +Bx+C=0$ . Solving for (5+X)(5-X)=7 by multiplying the factors and subtratcting 7 on both sides gives the equation $x^{2} -18=0$ , so
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6 0

3 years ago

Please help!!!!!!!!!!

defon

Answer: 90 + 4, third choice and 90 - -4, last choice

Her score is obviously higher when the mistake is fixed, adding the 4 points gives 94.

3 0

3 years ago

6. Describe a pattern shown in this sequence, and use the pattern to find the next two terms.

nlexa [21]

Answer:

The pattern is that each time, 8 is added. The next 2 numbers are 47, 55.

Step-by-step explanation:

7+8=15

15+8=23

23+8=31

31+8=39

39+8=47

47+8=55

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3 years ago

Find an exact value cos(19pi/12)

dmitriy555 [2]

$cos\left ( \frac{9\pi}{12}\right )=\frac{\sqrt{2-\sqrt{3}}}{2}$

Step-by-step explanation:

We know that

cos 2x = 2cos²x - 1

So we have

$cosx=\sqrt{\frac{1+cos2x}{2}}$

Substituting $x=\frac{19\pi}{12}$

$cos\left ( \frac{19\pi}{12}\right )=\sqrt{\frac{1+cos2\left ( \frac{19\pi}{12}\right )}{2}}\\\\cos\left ( \frac{19\pi}{12}\right )=\sqrt{\frac{1+cos\left ( \frac{19\pi}{6}\right )}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{1+cos570}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\\\cos\left ( \frac{9\pi}{12}\right )=\sqrt{\frac{2-\sqrt{3}}{4}}\\\\cos\left ( \frac{9\pi}{12}\right )=\frac{\sqrt{2-\sqrt{3}}}{2}$

$cos\left ( \frac{9\pi}{12}\right )=\frac{\sqrt{2-\sqrt{3}}}{2}$

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3 years ago

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