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3 years ago

How to write 20, 484, 163 and expanded form

1 answer:

7 0

20,484,163 in expanded form

2 ten million and 0 one million and 4 hundred thousand and 8 ten thousand and 4 one thousand and 1 hundreds and 6 tens and 3 ones.

or

2 x 10,000,000 + 0 x 1,000,000 + 4 x 100,000 + 8 x 10,000 + 4 x 1,000 + 1 x 100 + 6 x 10 + 3 x 1

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Sarah needs 30 liters of a 25% acid solution how many liters of the 10% and the 30% acid solutions should she mix to get what sh

sweet-ann [11.9K]

Answer:

7.5 L of 10% solution and 22.5 L of 30% solution

Step-by-step explanation:

Volume of 10% solution plus volume of 30% solution = total volume of 25% volume.

x + y = 30

Acid in 10% solution plus acid in 30% solution = total acid in 25% solution.

0.10 x + 0.30 y = 30 × 0.25

0.10 x + 0.30 y = 7.5

Solve the system of equations, using either substitution or elimination. I'll use substitution:

x = 30 − y

0.10 (30 − y) + 0.30 y = 7.5

3 − 0.10 y + 0.30 y = 7.5

0.20 y = 4.5

y = 22.5

x = 30 − y

x = 7.5

Sarah needs 7.5 L of 10% solution and 22.5 L of 30% solution.

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3 years ago

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Billy and Sam started with the same number of baseball cards in their collections. Billy collected 3 cards per week and now has

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Answer:

x+3y=29

x+2y=20

Step-by-step explanation:

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3 years ago

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Determine the absolute value of 5.<br> A) -5 <br> B) 0 <br> C) 10 <br> D) 5

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One household is to be selected at random from a town. The probability that the household has a cat is 0.20.2 . The proba

dimulka [17.4K]

Answer:

There is a 50% probability that the household has a dog, given that the household has a cat.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a household has a cat.

B is the probability that a household has a dog.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a household has a cat but not a dog and $A \cap B$ is the probability that a household has both a cat and a dog.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that the household has a cat or a dog is 0.5

$a + b + (A \cap B) = 0.5$

The probability that the household has a dog is 0.4

$B = 0.4$

$B = b + (A \cap B)$

$b = 0.4 - (A \cap B)$

The probability that the household has a cat is 0.2.

$A = 0.2$

$A = a + (A \cap B)$

$a = 0.2 - (A \cap B)$

$a + b + (A \cap B) = 0.5$

$0.2 - (A \cap B) + 0.4 - (A \cap B) + (A \cap B) = 0.5$

$A \cap B = 0.1$

What is the probability that the household has a dog, given that the household has a cat?

20% of the households have a cat, and 10% have both a cat and a dog. So

$P = \frac{A \cap B}{A} = {0.1}{0.2} = 0.5$

There is a 50% probability that the household has a dog, given that the household has a cat.

4 0

3 years ago

Pls help me I'll give brainlest and points

dmitriy555 [2]

Answer:

See below.

Step-by-step explanation:

O is the origin, so its abscissa is 0. O: 0

The only point 4 units from O is A, so A: 4

Point C has a positive abscissa like A.

Points B and C have opposite abscissas. The only opposites are -3 and 3, but C is positive, and B is negative.

B: -3

C: 3

D: -4.5

E: -6

M = (4 + (-6))/2 = -1. M: -1

OB = |0 - (-3)| = |3| = 3

DA = |-4.5 - 4| = |-8.5| = 8.5

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3 years ago