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2 years ago

What is 2.58333333333 as a fraction?

Mathematics

2 answers:

liberstina [14]2 years ago

8 0

Answer:

2 583/1000

Step-by-step explanation:

Kaylis [27]2 years ago

7 0

Answer

To write 2.58333333333 as a fraction you have to write 2.58333333333 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.

2.58333333333 = 2.58333333333/1 = 25.8333333333/10 = 258.333333333/100 = 2583.33333333/1000 = 25833.3333333/10000 = 258333.333333/100000 = 2583333.33333/1000000 = 25833333.3333/10000000 = 258333333.333/100000000 = 2583333333.33/1000000000 = 25833333333.3/10000000000 = 258333333333/100000000000

And finally we have:

2.58333333333 as a fraction equals 258333333333/100000000000

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Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a)

UNO [17]

Answer:

a) P=0.226

b) P=0.6

c) P=0.0008

d) P=0.74

Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

a) We calculate the probability that are 3 red, 2 blue, and 2 green balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_3^{12}\cdot C_2^{16}\cdot C_2^{18}=660\cdot 120\cdot 153=12117600$

Therefore, the probability is

$P=\frac{12117600}{53524680}\\\\P=0.226$

b) We calculate the probability that are at least 2 red balls.

We calculate the probability withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations: for 1 red balls

$C_1^{12}\cdot C_7^{34}=12\cdot 1344904=16138848$

Therefore, the probability is

$P_1=\frac{16138848}{53524680}\\\\P_1=0.3$

We calculate the number of favorable combinations: for none red balls

$C_7^{34}=5379616$

Therefore, the probability is

$P_0=\frac{5379616}{53524680}\\\\P_0=0.1$

Therefore, the the probability that are at least 2 red balls is

$P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6$

c) We calculate the probability that are all withdrawn balls are the same color.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056$

Therefore, the probability is

$P=\frac{44056}{53524680}\\\\P=0.0008$

d) We calculate the probability that are either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Let X, event that exactly 3 red balls selected.

$P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\$

Let Y, event that exactly 3 blue balls selected.

$P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\$

We have

$P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12$

Therefore, we get

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74$

8 0

3 years ago

4x - 7 = -3x<br><br> What is the answer?

soldi70 [24.7K]

Answer:

x = 1

Step-by-step explanation:

1. Add 3x to both sides of the problem

4x + 3x - 7 = -3x + 3x
-3x + 3x cancels out and leaves us with zero
7x - 7 = 0

2. Add 7 to both sides

7x (- 7 + 7) = 0 + 7
-7 + 7 cancels out
7x = 7

3. Divide both sides by 7

7x/7 = 7/7
x = 1

To make sure our answer is correct we can plug in 1 to x in the problem

4(1) - 7 = -3(1)
4 - 7 = -3 : True

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