Mathematics, 31.08.2019 14:30 alvis4485

Afair, six-sided die (numbered 1 to 6) is rolled eight times to form an eight-digit number. what is the probability that the resulting number is a multiple of 8? express your answer as a common fraction.

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A number is a multiple of 8 if the last 3 digits of it, form a number which is a multiple of 8.

For example: 45288, 16840, 90024 are all multiples of 8, because 288, 840 and 024, the last 2 digits, are multiples of 8.

The total number of 8-digit numbers that we can form using {1,2,3,4,5,6} is $6*6*6*6*6*6*6*6= 6^{8}$ , as we have 6 options for each digit.

The last 3 digits of multiple of 8, formed from the set {1,2,3,4,5,6} can be one of the following:

{112, 136, 144, 152, 216, 224,232, 256, 264
312, 336, 344, 352, 416, 424,432, 456, 464
512, 536, 544, 552, 616, 624,632, 656, 664} ,

which is 27 of them.

For each of these 3 digits fixed in the end, we can form $6*6*6*6*6*1*1*1= 6^{5}$ numbers

so in total, there are $27*6^{5}$ numbers divisible by 8, whose digits are numbers in the set {1, 2, 3, 4, 5, 6}.

P(multiple of 8)=n(multiple of 8)/n(numbers that can be formed)=

$\frac{27*6^{5}}{6^{8}} = \frac{27}{6^{3}}= ( \frac{3}{6} )^{3} =( \frac{1}{2} )^{3}=0.125$

0.125

Answer from: Quest

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.

Answer from: Quest

$3 \times 4 = 12 \\ 6 \times 8 = 48$ $48 + 12 = 60$ hope it is correct if not i'm srry i tried to

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Afair, six-sided die (numbered 1 to 6) is rolled eight times to form an eight-digit number. what is...

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