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3 years ago

A2. Find the rational number whose decimal expansion is 0.3344444...

Mathematics

2 answers:

Vladimir79 [104]3 years ago

7 0

Answer:

$\large\boxed{0.334444...=\dfrac{301}{900}}$

Step-by-step explanation:

$x=0.334444...\qquad\text{multiply both sides by 100}\\\\100x=33.4444...\qquad\text{multiply both sides by 10}\\\\1000x=334.4444...\qquad\text{subtract}\ 100x\ \text{from}\ 1000x\\\\1000x-100x=334.4444...-33.4444...\\\\900x=301\qquad\text{divide both sides by 900}\\\\x=\dfrac{301}{900}$

JulsSmile [24]3 years ago

5 0

Answer:

301/900.

Step-by-step explanation:

0.33 = 33/100

44.44444,, = 44/99

So 0.004444...

= 44/9900

33/100 + 44/9900

301/900

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Pleaseee pleaseee help 10 mins left

Lesechka [4]

Answer:

fraction is 44/100

decimal form is 0.44

percentage is 44%

Step-by-step explanation:

follow me for my effort

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3 years ago

40 POINTSSSSSS> RIGHT ANSWER ONLY!!!!. Will report if answer is faulty. PLEASE HELP MATHEMATICIANS!!!!!!!

Marat540 [252]

The probability of hitting the black circle is closer to 0 than it is to 1. because: find area of circle ~ 3.14*r^2
which is 3.14 because the radius squared is 1. so 1*3.14= 3.14
and the area of the whiteness is 121.
subtract 3.14 from 121 and you get 117.86
this answer is closer to a 0 probability than the white space. the white space is closer to a probability of 1.
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7 0

4 years ago

You toss three 6-sided dice and record the sum of the three faces facing up. a) Describe precisely a sample space S for this exp

timama [110]

Answer:

a.)Sample space means all possible outcomes, since we know the dice are 6 face, then the sample space becomes all possible outcomes when we toss the die.

b.) 9/216

c.) 9/216

d.) 212/216

Step-by-step explanation:

Sample space means all possible outcomes, since we know the dice are 6 face, then the sample space becomes all possible outcomes when we toss the die.

[1,1,1] [1,1,2] [1,1,3] [1,1,4] [1,1,5] [1,1,6]

[1,2,1] [1,2,2] [1,2,3] [1,2,4] [1,2,5] [1,2,6],

[1,3,1] [1,3,2] [1,3,3] [1,3,4] [1,3,5] [1,3,6]

[1,4,1] [1,4,2] [1,4,3] [1,4,4] [1,4,5] [1,4,6]

[1,5,1] [1,5,2] [1,5,3] [1,5,4] [1,5,5] [1,5,6]

[1,6,1] [1,6,2] [1,6,3] [1,6,4] [1,6,5] [1,6,6]

[2,1,1] [2,1,2] [2,1,3] [2,1,4] [2,1,5] [2,1,6]

[2,2,1] [2,2,2] [2,2,3] [2,2,4] [2,2,5] [2,2,6],

[2,3,1] [2,3,2] [2,3,3] [2,3,4] [2,3,5] [2,3,6]

[2,4,1] [2,4,2] [2,4,3] [2,4,4] [2,4,5] [2,4,6]

[2,5,1] [2,5,2] [2,5,3] [2,5,4] [2,5,5] [2,5,6]

[2,6,1] [2,6,2] [2,6,3] [2,6,4] [2,6,5] [2,6,6]

[3,1,1] [3,1,2] [3,1,3] [3,1,4] [3,1,5] [3,1,6]

[3,2,1] [3,2,2] [3,2,3] [3,2,4] [3,2,5] [3,2,6],

[3,3,1] [3,3,2] [3,3,3] [3,3,4] [3,3,5] [3,3,6]

[3,4,1] [3,4,2] [3,4,3] [3,4,4] [3,4,5] [3,4,6]

[3,5,1] [3,5,2] [3,5,3] [3,5,4] [3,5,5] [3,5,6]

[3,6,1] [3,6,2] [3,6,3] [3,6,4] [3,6,5] [3,6,6]

[4,1,1] [4,1,2] [4,1,3] [4,1,4] [4,1,5] [4,1,6]

[4,2,1] [4,2,2] [4,2,3] [4,2,4] [4,2,5] [4,2,6],

[4,3,1] [4,3,2] [4,3,3] [4,3,4] [4,3,5] [4,3,6]

[4,4,1] [4,4,2] [4,4,3] [4,4,4] [4,4,5] [4,4,6]

[4,5,1] [4,5,2] [4,5,3] [4,5,4] [4,5,5] [4,5,6]

[4,6,1] [4,6,2] [4,6,3] [4,6,4] [4,6,5] [4,6,6]

[5,1,1] [5,1,2] [5,1,3] [5,1,4] [5,1,5] [5,1,6]

[5,2,1] [5,2,2] [5,2,3] [5,2,4] [5,2,5] [5,2,6],

[5,3,1] [5,3,2] [5,3,3] [5,3,4] [5,3,5] [5,3,6]

[5,4,1] [5,4,2] [5,4,3] [5,4,4] [5,4,5] [5,4,6]

[5,5,1] [5,5,2] [5,5,3] [5,5,4] [5,5,5] [5,5,6]

[5,6,1] [5,6,2] [5,6,3] [5,6,4] [5,6,5] [5,6,6]

[6,1,1] [6,1,2] [6,1,3] [6,1,4] [6,1,5] [6,1,6]

[6,2,1] [6,2,2] [6,2,3] [6,2,4] [6,2,5] [6,2,6],

[6,3,1] [6,3,2] [6,3,3] [6,3,4] [6,3,5] [6,3,6]

[6,4,1] [6,4,2] [6,4,3] [6,4,4] [6,4,5] [6,4,6]

[6,5,1] [6,5,2] [6,5,3] [6,5,4] [6,5,5] [6,5,6]

[6,6,1] [6,6,2] [6,6,3] [6,6,4] [6,6,5] [6,6,6]

b.) Probability that the sum is 16 or more is

Pr[4,6,6] + pr[5,5,6] + pr [ 5,6,5] + pr [5,6,6] + pr [6,5,5] + pr [6,5,6] + pr [6,6,4] + pr[6,6,5] + pr [6,6,6]

Becomes:

[1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 9/216

Probability that the sum is 4 or 5 is

Pr [ 1,1,2] or pr[1,2,2] or pr [1,1,3] or pr [1,2,1] or pr[2,1,2] or pr[1,3,1] or pr[3,1,1] or pr[2,1,1] or pr[2,2,1]

Becomes:

[1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 9/216

Probability that the sum is less than 17

We take it as:

1- probability that the sum is 17 and above.

Now probability that the sum is 17 and above becomes

pr[5,6,6] or pr[6,5,6] or pr[6,6,5] or pr[6,6,6]

= [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 4/216

Hence, probability that the sum is less than 17 becomes:

1-4/216 = 212/216.

3 0

3 years ago

Plz I need help with this I am struggling failing math

Goryan [66]

Answer:

CB = 5, $m\angle{BDA} = 23^\circ$

Step-by-step explanation:

Assuming that the marks next to angle D mean that $\angle{BDA} \cong \angle{BDC}$ , then there is enough information to conclude that the two right triangles are congruent (HA theorem--hypotenuse/angle).