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Gnesinka [82]
3 years ago
7

Fracions closer to 1 than 0

Mathematics
2 answers:
defon3 years ago
7 0

1/2 (0.5), 2/3 (0.66), 3/4 (0.75) are all fractions closer to 1 than to 0.


Hope it helped,

Happy homework/ study/ exam!

ruslelena [56]3 years ago
5 0
1 1/3, 1 1/4, 1 1/5 .....

Please vote my answer brainliest. thanks!
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makvit [3.9K]

Answer:

b

Step-by-step explanation:

5 0
2 years ago
Percents that are added together are _______ percents.
vivado [14]
Percents that are added together are combining percents.
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3 years ago
Find the larger of two consecutive integers if the larger is 149 less than twice the smaller.
snow_tiger [21]

Answer:

151

Step-by-step explanation:

Let x = the smaller integer and y = the larger integer.

We have two conditions:

(1)   x + 1 = y

(2)       y =  2x - 149      Substitute (1) into (2)

    x + 1 = 2x - 149        Add 150 to each side

x + 150 = 2x                 Subtract x from each side

        x = 150                Substitute into (1)

150 +1 = y

       y = 151

The larger integer is 151.

<em>Check: </em>

(1) 150 + 1 = 151

          151 = 151

(2)       151 = 2×150 -149

          151 = 300 -149

          151 = 151

4 0
3 years ago
Please help me (geometry)
vesna_86 [32]

Answer:

True

Step-by-step explanation:

When two lines are parallel, the corresponding angles formed by the transversal (intersecting line) are congruent. This is called the corresponding angles postulate. It can be used in reverse: if two corresponding angles are congruent, then the non-transversal lines are parallel.

4 0
3 years ago
Read 2 more answers
A bag contains eleven equally sized marbles, which are numbered. Two marbles are chosen at random and replaced after each select
Digiron [165]

Answer:

B. \frac{24}{121}.

Step-by-step explanation:

Total number of marbles = 11.

Since, after choosing the first marble, we are putting it back and then the second marble is chosen.

As, there are 4 shaded marbles.

So, the probability of getting the first marble shaded = \frac{\binom{4}{1}}{\binom{11}{1}} = \frac{4}{11}

Also, there are 6 odd labeled marbles.

So, the probability of getting the second marble being odd labeled = \frac{\binom{6}{1}}{\binom{11}{1}} = \frac{6}{11}

So, the probability of getting the first marble shaded and second marble labeled odd = \frac{4}{11}\times \frac{6}{11} = \frac{24}{121}.

Hence, the required probability is \frac{24}{121}.

4 0
3 years ago
Read 2 more answers
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