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ExtremeBDS [4]
3 years ago
14

Which set of integers does NOT represent the lengths of the sides of a triangle? A. {6,6,11} B. {9,10,11} C. {4,8,12} D. {4,7,9}

Mathematics
1 answer:
Vladimir79 [104]3 years ago
3 0

Answer:

C

Step-by-step explanation:

I suppose you have learned that for the sides of a triangle to work, it has to be a + b > c, the 4 is the a, the 8 is the b, the 12 is the c.

So: 4 + 8 > 12; however this is not true, they are equal so the triangle wont be a triangle, it would be lines that never connect.

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A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa
gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4  | red dice) = \dfrac{1}{6}

P (4 | green dice) = \dfrac{3}{6} =\dfrac{1}{2}

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = \dfrac{1}{2}

The probability of two 1's and two 4's in the first dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^4

= \dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4

= 6 \times ( \dfrac{1}{6})^4

= (\dfrac{1}{6})^3

= \dfrac{1}{216}

The probability of two 1's and two 4's in the second  dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^2  \times  \begin {pmatrix} \dfrac{3}{6}  \end {pmatrix}  ^2

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= 6 \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= ( \dfrac{1}{6}) \times  ( \dfrac{3}{6})^2

= \dfrac{9}{216}

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = \dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}

The probability of two 1's and two 4's in both die = \dfrac{1}{432}  + \dfrac{1}{48}

The probability of two 1's and two 4's in both die = \dfrac{5}{216}

By applying  Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's )  = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's )  = \dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}

P(second die (green) | two 1's and two 4's )  = \dfrac{0.5 \times 0.04166666667}{0.02314814815}

P(second die (green) | two 1's and two 4's )  = 0.9

Thus; the probability the die chosen was green is 0.9

8 0
3 years ago
Find the number C(10,6)
EastWind [94]

Answer:

210.

Step-by-step explanation:

Combination formula is

nCr = \frac{n!}{r!(n-r)!}

Then, we have that n = 10 and r=6:

C(10,6) = \frac{10!}{6!(10-6)!}

C(10,6) = \frac{10!}{6!4!}

To simplify calculus, we are going to use that n! = (n-1)!n = (n-2)!(n-1)n and so on.

C(10,6) = \frac{6!*7*8*9*10}{6!4!}

C(10,6) = \frac{7*8*9*10}{4!}

C(10,6) = \frac{5040}{4*3*2*1}

C(10,6) = \frac{5040}{24}

C(10,6) = 210.

5 0
2 years ago
This is homework. please help with the answer for both 11a and 11b :))
RideAnS [48]
11a:
3x+5=5x-57
3x-5x=-57-5
-2x=-62
x=31
11b:
2x+2x+4x+150+4x+150=360
12x+300=360
12x=360-300
12x=60
x=5

hope this helped !!
3 0
2 years ago
I don’t get it or understand
frutty [35]

Answer:

$1.30

Step-by-step explanation:

What you do is add the original 65 cents to the additional 65 cents that were shown to get 130. Since it's more than 100 you make it into dollars, so the final answer is $1.30.

6 0
2 years ago
Read 2 more answers
The length of the base of a triangle is twice it’s height. If the area of the triangle is 441 square kilometers, find the height
Sonja [21]

Answer:

21 kilometers

Step-by-step explanation:

Let the height be x. Then, the length of the base is 2x. The formula for the area is of the triangle is given by base*height/2. Therefore, the area of the triangle is equal to \frac{x \cdot 2x}{2} = x^2, which is in turn equal to 441. Since x must be positive, then 21^2=441, meaning that the height is 21 kilometers.

5 0
2 years ago
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