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3 years ago

Write the number21,989.098 in expanded form and word form

Mathematics

1 answer:

DiKsa [7]3 years ago

4 0

Expanded Form:
20,000+1,000+900+80+9+0.09+0.008

Word Form:

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How to solve for x: 2x+y=3 and x+y=9

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The answer for the first one would be: y=3-2x
The answer for the second one would be: y=9-x

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3 years ago

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Find the scale factor of the smaller figure to the larger figure

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(smaller length measure) to (larger length measure)
3 to 9 or 6 to 18
reduces to 1 to 3 or $\frac{1}{3}$

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3 years ago

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Write an equation of the line passing through the points (6, -8) and (1, -3)

Helga [31]

Answer:

Step-by-step explanation:

(-3+8)/(1-6)= 5/-5 = -1

y + 8 = -(x - 6)

y + 8 = -x + 6

y = -x - 2

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3 years ago

Kathy's square garden has an area of 105 square feet. How long will one side of a fence around the garden be?

Stels [109]

Because the garden is a square, each side would be the same length. You take the square root of 105. It is 10.246
Answer: between 10 and 11

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4 years ago

A deck of 52 cards contains 12 picture cards. If the 52 cards are distributed in a random manner among four players in such a wa

Mkey [24]

Answer:

The probability that each player will receive three picture cards = 0.0324

Step-by-step explanation:

As given,

A deck of 52 cards contains 12 picture cards

Remaining card = 52 - 12 = 40

So,

Total number of ways in which 12 picture card is distributed = $\frac{12!}{3! 3! 3! 3!}$

Now,

The Total number of ways in which Remaining cards are distributed = $\frac{40!}{10! 10! 10! 10!}$

So,

Total number of ways of getting 3 picture card and remaining card = $\frac{12!}{3! 3! 3! 3!}$ × $\frac{40!}{10! 10! 10! 10!}$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$

Now,

Total number of ways to distribute 52 cards so that each people get 13 card = $\frac{52!}{13! 13! 13! 13!} = \frac{52!}{ (13!)^{4} }$

∴ The probability = $\frac{\frac{12! 40!}{(3!)^{4} (10!)^{4} }}{\frac{52!}{(13!)^{4} }}$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$ × $\frac{(13!)^{4} }{ 52! }$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$ × $\frac{(13.12.11.10!)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41.40! }$

= $\frac{12!}{(3!)^{4} }$ × $\frac{(13.12.11)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}$

= $\frac{479,001,600}{(6)^{4} }$ × $\frac{(1716)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}$

= 0.0324

∴ we get

The probability that each player will receive three picture cards = 0.0324

6 0

3 years ago