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3 years ago

13

Plz someone help me on this too plz plz

1 answer:

san4es73 [151]3 years ago

8 0

1. 7 and -7 are both 7 units away from 0. They are 14 units away from each other. Absolute value helps because rather than 7+(-7) being 0 absolute value makes anything positive so it becomes 7+7=14 to be able to count the units between the two.

2. 5 and -5 are both 5 units away from 0. They are 10 units away from each other. Absolute value helps because rather than 5+(-5) being 0 absolute value makes anything positive so it becomes 5+5=10 to be able to count the units between the two.

3. 2 and -2 are both 2 units away from 0. They are 4 units away from each other. Absolute value helps because rather than 2+(-2) being 0 absolute value makes anything positive so it becomes 2+2=4 to be able to count the units between the two.

Hope that helps

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If one bag of topsoil covers 10 square feet, how many bags are needed to cover this flowr bed?

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3 years ago

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I am having trouble with number 2. Please show how the answer was gotten.

Andrews [41]

We know how we got 90, since a right angle is 90 degrees,

180 degrees are in a triangle.

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4 0

3 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

(w+3/w+4)=(w-1/w+7)+1

RoseWind [281]

I hope this helps you

5 0

3 years ago