Mathematics, 28.01.2020 21:47 aaamanda99

Afair 6-faced die is tossed twice. letting e and f represent the outcomes of the each toss (which are independent), compute the following probabilities. a. the sum of e and f is 11 b. the sum of e and f is even c. the sum of e and f is odd and the sum is greater than 3 d. e is even and less than 6, and f is odd and greater than 1 e. e is greater than 2, and f is less than 4 f. e is 4, and the sum of e and f is odd

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Answer from: biggiecheese93

(a) $\frac{1}{18}$ (d) $\frac{1}{9}$

(b) $\frac{1}{2}$ (e) $\frac{1}{3}$

Step-by-step explanation:

On tossing a 6-face die twice the outcomes of E and F are:

{1, 2, 3, 4, 5 and 6}

And there are total 36 outcomes of the form (E, F).

(a)

The sample space of getting a sum of 11 is: {(5, 6) and (6, 5)}

The probability of getting a sum of 11 is:

$P(Sum 11) =\frac{2}{36} \\=\frac{1}{18}$

(b)

The sample space of getting an even sum is:

{(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6) (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6) (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}

The probability of getting an even sum is:

$P(Even Sum)=\frac{18}{36}\\ =\frac{1}{2}$

(c)

The sample space of getting an odd sum more than 3 is:

{(1, 4), (1, 6), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3) and (6, 5)}

The probability of getting an odd sum more than 3 is:

P (Odd Sum more than 3) =

$=\frac{16}{36}\\=\frac{4}{9}$

(d)

The sample space of (E, F) such that E is even and less than 6, and F is odd and greater than 1 is:

{(2, 3), (2, 5), (4, 3) and (4, 5)}

The probability such that E is even and less than 6, and F is odd and greater than 1 is:

$P(Even E1) =\frac{4}{36} \\=\frac{1}{9}$

(e)

The sample space of E and F such that E is more than 2, and F is less than 4 is:

E = {3, 4, 5 and 6} F = {1, 2 and 3}

Then the total outcomes of (E, F) will be 12.

The probability such that E is more than 2, and F is less than 4 is:

P(E2, F

(f)

The sample space of (E, F) such that E is 4 and sum of E and F is odd is:

{(4, 1), (4, 3) and (4, 5)}

The probability such that E is 4 and sum of E and F is odd is:

$P(E=4, E+F = Odd)=\frac{3}{36} \\=\frac{1}{12}$

Answer from: Quest

$w\geq 14$

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

$2l+2w\geq 40$

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

$2\times 6+2w\geq 40$

now, we can solve for w

$12+2w\geq 40$

subtract both sides by 12

$12+2w-12\geq 40-12$

$2w\geq 28$

divide both sides by 2

and we get

$w\geq 14$

so,

the smallest possible measurement of the width is 14 feet

Answer from: Quest

The key is the word about, they want an estimation.63 is closer to 60, 71 is closer to 7060 + 70 = 130ryan counted about 130 birds

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