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BARSIC [14]
3 years ago
14

You roll a fair 6 sided dice. What is p(not 5)?

Mathematics
1 answer:
liq [111]3 years ago
7 0

Answer:

p(not 5) = 5/6

Step-by-step explanation:

The probability that we roll a 5 is 1/6. So, the probability we don't roll a five is simple 1 minus that:

1-\frac{1}{6} = \frac{5}{6}

And that's your answer.

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Flura [38]

Answer:

280 cm

or 280cm^2

Step-by-step explanation:

8 0
3 years ago
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Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .
kogti [31]

Answer:

The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

b = \sqrt{1 + 2\cdot a \cdot c} (2)

c is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if (x+y)^{2} = x^{2} + 2\cdot x\cdot y  + y^{2}. From statement, we must fulfill the following identity:

a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}

By Associative and Commutative properties, we can reorganize the expression as follows:

a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2} (1)

Then, we have the following system of equations:

x = a (2)

(b^{2}-1) = 2\cdot x\cdot y (3)

y = c (4)

By (2) and (4) in (3), we have the following expression:

(b^{2} - 1) = 2\cdot a \cdot c

b^{2} = 1 + 2\cdot a \cdot c

b = \sqrt{1 + 2\cdot a\cdot c}

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, a, b, c > 1. If a, b and c are prime numbers, then  2\cdot a\cdot c must be an even composite number, which means that a and c can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, b must be a natural number, which means that:

1 + 2\cdot a\cdot c \ge 4

2\cdot a \cdot c \ge 3

a\cdot c \ge \frac{3}{2}

But the lowest possible product made by two prime numbers is 2^{2} = 4. Hence, a\cdot c \ge 4.

The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

b = \sqrt{1 + 2\cdot a \cdot c} (2)

c is another arbitrary prime number. (3)

Example: a = 2, c = 2

b = \sqrt{1 + 2\cdot (2)\cdot (2)}

b = 3

4 0
3 years ago
Answer question 8,9 and 10 with full details.
Lelechka [254]

Answer:

See Explanation

Step-by-step explanation:

8. (1)

Area of trapezoid = 1/2(sum of || sides) *perpendicular distance

36 =1/2(x+ y) * 6

36 = ((x + y)*3

36/3 = x + y

12 = x + y

x + y = 12.... (1)

8(2)

It is given that:

x = 2y....(2)

From equations (1) & (2)

2y + y = 12

3y = 12

y = 12/3

y = 4

From equation (2)

x = 2y = 2*4

x = 8

9.

Area of square = 1 square m

Side of square = 1 m

Area of circle = 1 square m

\pi {r}^{2}  = 1 \\  {r}^{2}  =  \frac{1}{\pi}  \\ r =  \frac{1}{ \sqrt{\pi} }  \\ c = 2\pi \: r \\ c = 2\pi \times  \frac{1}{ \sqrt{\pi} }  \\ c = 2 \sqrt{\pi}  \\ c = 2 \sqrt{3.14}  \\ c = 2 \times 1.77200451 \\ c = 3.54400902 \\ c = 3.544 \\ difference = 3.544 - 1 = 2.544  \\difference \approx \: 2.54

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3 years ago
12-3 (2w+1)=7w-3 (7+w)
n200080 [17]
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4r^2=4 if r=1
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3 years ago
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