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kirza4 [7]
3 years ago
6

A fair, twenty-faced die has $19$ of its faces numbered from $1$ through $19$ and has one blank face. another fair, twenty-faced

die has $19$ of its faces numbered from $1$ through $8$ and $10$ through $20$ and has one blank face. when the two dice are rolled, what is the probability that the sum of the two numbers facing up will be $24?$ express your answer as a common fraction.
Mathematics
1 answer:
lawyer [7]3 years ago
8 0
Represent 24 as a sum of two numbers, first number from first die and second number from second die.
24=19+5=18+6=17+7=16+8=14+10=13+11=12+12=11+13=10+14=9+15=8+16=7+17=6+18=5+19=4+20 (the sums 15+9 and 20+4 are absent, because there aren't numbers: 20 on the first die and number 9 on the second die). Totally, you receive 15 different representations of 24.
<span>The probability that the sum of the two numbers facing up will be 24 is
</span><span>
</span><span>P= \frac{15}{20\cdot 20} = \frac{3}{80} (here 20\cdot 20 means that you have 20 possibilities to roll first number or blank face on the first die and 20 possibilities to roll number or blank face on the second die).
</span>


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-x + 2y = 3<br> 2x – 3y = -6
s2008m [1.1K]

Answer:

x = -3

y = 0

Step-by-step explanation:

<u>Given</u><u> </u><u>equations</u><u> </u><u>:</u><u>-</u><u> </u>

<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u>

<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u>

<u>From</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u><u> </u><u> </u>

<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u>

<u>-x</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>-</u><u> </u><u>2</u><u>y</u><u> </u>

<u>x</u><u> </u><u>=</u><u> </u><u>2</u><u>y</u><u> </u><u>-</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u>

<u>From</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u><u> </u>

<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u>

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<u>x =  \frac{ - 6 + 3y}{2}</u>

<u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>

<u>Equating</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u><u> </u><u>and</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>

<u>x</u><u> </u><u>=</u><u> </u><u>x</u><u> </u>

<u>2y - 3 =  \frac{ - 6 + 3y}{2}</u>

4y - 6 = -6 + 3y

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Putting value of y in ( iii )

x = 2y - 3

x = 2 ( 0 ) - 3

x = -3

4 0
3 years ago
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Mkey [24]
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4 0
3 years ago
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nadya68 [22]

Answer:

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Explanation:

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8 0
3 years ago
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babunello [35]

yes it is correct

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47/6=47/6

7 0
2 years ago
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