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3 years ago

Simplify the ratio 4:22:30

2 answers:

5 0

4:22:30
you have to look for the number which can divide all 3 of them
which in this case is 2

divide by 2

4/2 : 22/2 : 30/2

2:11:15

now there is no number which can divide all of them so
2:11:15 is the answer

pshichka [43]3 years ago

4 0

Relation Extremes equal means.
4×30÷22.
Or
(22)*2=4×30.=120.
484=120
484/120

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The total length of pencils A, B and C is 29 cm. Pencil a is 11 cm shorter then pencil B, and pencil B is twice as long a pencil

satela [25.4K]

The length of pencil A is 5 cm

Solution:

Let the length of pencil A be "x"

Let the length of pencil B be "y"

Let the length of pencil C be "z"

The total length of pencils A, B and C is 29 cm

Therefore,

length of pencil A + length of pencil B + length of pencil C = 29

x + y + z = 29 ------------ eqn 1

Pencil A is 11 cm shorter then pencil B

x = y - 11 ------- eqn 2

Pencil B is twice as long a pencil C

y = 2z

$z = \frac{y}{2}$ ------ eqn 3

Substitute eqn 2 and eqn 3 in eqn 1

$y - 11 + y + \frac{y}{2} = 29\\\\2y + \frac{y}{2} = 29 + 11\\\\\frac{4y+y}{2} = 40\\\\5y = 80\\\\y = 16$

Substitute y = 16 in eqn 2

x = 16 - 11

x = 5

Thus length of pencil A is 5 cm

5 0

3 years ago

You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob

Ilia_Sergeevich [38]

Answer:

a) No

b) 42%

c) 8%

d) X 0 1 2

P(X) 42% 50% 8%

e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6

P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

c) P(win first game) = 0.4

P(win second game) = 0.2

P(win both games) = P(win first game) × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X 0 1 2

P(X) 42% 50% 8%

P(X = 0) = P(lose both games) = P(lose first game) × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game) × P(win second game)] + [ P(win first game) × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value $\mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66$