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3 years ago

Which fraction is the largest?

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

8 0

Answer:

5/6

Step-by-step explanation:

1/3 has two more thirds until it's a whole

17/20 has three more twentieths until it's a whole

1/4 has four more fourths until it's a whole

5/6 has <u>one</u> more sixth until it's a whole

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Jose has -75 points. He wins 200 points;then he loses 25 points. How many points does he have now?

Sav [38]

Answer:

100 points

Step-by-step explanation:

-75 + 200 = 125

125 - 25 = 100

3 0

2 years ago

Read 2 more answers

Determine if its possible to form a triangle using the set of segments with the given measurement. explain please! 12 m, 5 m, 18

IRINA_888 [86]

Adding together the shorter two sides results in 17 m. If these two sides were laid end to end, the sum of their sides (17 m) would be less than the length of the third side. Thus, NO triangle could be constructed with this set of line segments.

6 0

3 years ago

How do I solve this?

Aleksandr-060686 [28]

First you have to set up your equation. 5x+80 and x+15.

8 0

3 years ago

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Please help me!!!!!!

goldfiish [28.3K]

Answer:

False, there are no solutions

Step-by-step explanation:

y = 2x+4

y = 2x+10

The slopes are the same (2) but the y intercepts are different

These lines are parallel with different intercepts

They will never intersect so there are no solutions

3 0

2 years ago

Read 2 more answers

Consider two people being randomly selected. (For simplicity, ignore leap years.)

inna [77]

Answer:

$(a) = \frac{144}{133225} \\\\(b) = \frac{1}{365}$

Step-by-step explanation:

Part (a) the probability that two people have a birthday on the 9th of any month.

Neglecting leap year, there are 365 days in a year.

There are 12 possible 9th in months that make a year calendar.

If two people have birthday on 9th; P(1st person) and P(2nd person).

$=\frac{12}{365} X\frac{12}{365} = \frac{144}{133225}$

Part (b) the probability that two people have a birthday on the same day of the same month

P(2 people selected have birthday on the same day of same month) + P(2 people selected not having birthday on same day of same month) = 1

P(2 people selected not having birthday on same day of same month):

$= \frac{365}{365} X \frac{364}{365} =\frac{364}{365}$

P(2 people selected have birthday on the same day of same month) $= 1-\frac{364}{365} \\\\= \frac{1}{365}$

7 0

3 years ago