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allochka39001 [22]
3 years ago
14

While Playing Yahtzee and rolling 5 dice, Mei gets the result shown at right (the numbers are 2, 4, 4, 4, and 1). Mei decides to

keep the three fours and reroll the other 2 dice. a. what is the probability that Mei will have 5 of a kind? b. What is the probability that she will have 4 of a kind? c. what is the probability that she will have exactly three 4's?
Mathematics
1 answer:
Elina [12.6K]3 years ago
5 0
A. The probability is 1/36. There are 36 combinations of the first and second dice and only one of them is 2 - 4s.

B. 10/36. Ten of the options contain exactly 1 - 4. 

C. The remaining 25/36 options do not include a 4. 
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4 - The area of a rectangle is 256.5m. If the length is 18m what is the perimeter of the rectangle? -
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The equation of the area of a triangle is Area=Length×Width. We know area=256.5 and length is 18. So now we know 256.5=18×width. We then divide 256.5÷18 to get the width. The width is 14.25. Perimeter=legth+length+width+width. If width is 18 and length is 14.25 then we add up all 4 sides. 18+18+14.25+14.25=64.5. 64.5 is the perimeter
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Help please I'm to stupid for this​
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Answer:

X>0, so the second one

Step-by-step explanation:

Since the point on 0 is OPEN that means the number (0) is NOT part of the solution. Therefore, X is greater than 0.

4 0
3 years ago
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givi [52]

Given: (a) 2(x+3)=2x+6 and (b) 4(y-3)=4y-12.

To find: The expanded form of the given expressions.

(c) 4(m+n)=4m+4n       (using a(b+c)=ab+ac)

(d) 3(5-q)=15-3q           (using a(b-c)=ab-ac)

(e) 5(2c+1)=10c+5          (using a(b+c)=ab+ac)

(f) 3(2x-5)=6x-15          (using a(b-c)=ab-ac)

(g) 7(4b-1)=28b-7          (using a(b-c)=ab-ac)

(h) 3(2x+y-5)=3((2x+y)-5)

⇒3(2x+y-5)=3(2x+y)-15         (using a(b-c)=ab-ac)

⇒3(2x+y-5)=6x+3y-15            (using a(b+c)=ab+ac)

(i) 2(6a-4b+3)=2((6a-4b)+3)

⇒2(6a-4b+3)=2(6a-4b)+6         (using a(b+c)=ab+ac)

⇒2(6a-4b+3)=12a-8b+6            (using a(b-c)=ab-ac)

(j)6(m+n+p)=6((m+n)+p)

⇒ 6(m+n+p)=6(m+n)+6p           (using a(b+c)=ab+ac)

⇒ 6(m+n+p)=6m+6n+6p            (using a(b+c)=ab+ac)

(k) y(y+2)=y^2+2y          (using a(b+c)=ab+ac)

(l) g(g-3)=g^2-3g           (using a(b-c)=ab-ac)

(m) n(4-n)=4n-n^2        (using a(b-c)=ab-ac)

(n) a(b+c)=ab+ac           (using a(b+c)=ab+ac)

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(p) 2x(x+5)=2x^2+10x     (using a(b+c)=ab+ac)

(q) 4y(x-3)=4xy-12y      (using a(b-c)=ab-ac)

(r) 5a(2b-5)=10ab-25a     (using a(b-c)=ab-ac)

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(t) 5p(4p-5q)=20p^2-25pq    (using a(b-c)=ab-ac)

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