.0476190476 as a fraction

1 answer:

3 0

Step 1: Rewrite the decimal number as a fraction with 1 in the denominator.
Step 2: Multiply to remove 10 decimal places. Here, you multiply top and bottom by 10 to the power of 10 = 10000000000
Step 3: Find the GCF of 476190476 and 10000000000 if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 4.
Step 4: denominator by GCF = 4,476190476÷410000000000÷4=1190476192500000000476190476÷410000000000÷4=1190476192500000000ThereforeX=119047619/2500000000X=1190476192500000000In conclusion,.0476190476=119047619/2500000000

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. The mean width of 12 iPads is 5.1 inches. The mean width of 8 Kindles is 4.8 inches. a. What is the total width of the iPads?

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Answer:

a) 61.2, b) 38.4 and c) 4.98

Step-by-step explanation:

Given:

The mean width of 12 I-Pads is 5.1 inches.

The mean width of 8 Kindles is 4.8 inches.

Question asked:

a. What is the total width of the I-Pads?

b. What is the total width of the Kindles?

c. What is the mean width of the 12 I-Pads and 8 Kindles?

Solution:

As we know:

$Mean =\frac{Sum\ of\ observations}{Number\ of\ observations}\\ \\ Mean\ width =\frac{Sum\ of\ width\ of\ all \ I-pad}{Number \ of\ I-pad}$

$5.1=\frac{Sum\ of\ width\ of\ all\ I-pad}{12} \\ \\ By \ cross\ multiplication\\ \\ Sum\ of\ width\ of\ all\ I-pad}=5.1\times1 2\\ \\ Sum\ of\ width\ of\ all\ I-pad}= 61.2$

a) Thus, the total width of the I-Pads are 61.2 inches.

$Mean\ width =\frac{Sum\ of\ width\ of\ all\ Kindles}{Number \ of\ Kindles}$

$4.8=\frac{Sum\ of\ width\ of\ all\Kindles}{8} \\ \\ By \ cross\ multiplication\\ \\ Sum\ of\ width\ of\ all \ Kindles=4.8\times8\\ \\ Sum\ of\ width\ of \ all\ Kindles}=38.4$