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Hunter-Best [27]
4 years ago
10

A game designer must decide how to color five buildings that are in a row. Using only the colors yellow, green, red, and blue, e

ach building must be painted with exactly one color. Any two neighboring buildings must be different colors, and the first, third, and fifth buildings must be different colors. How many ways are there to paint the five buildings?
Mathematics
1 answer:
saw5 [17]4 years ago
3 0

Answer:

A million

Step-by-step explanation:

?

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I need help with this question. use absolute value<br><br>9 = |X+5|​
Reptile [31]

Answer:

x = -14, 4

Step-by-step explanation:

9 = |x + 5|

~Split into 2 equations

x + 5 = -9 and x + 5 = 9

~Solve for x in both

x + 5 = -9

x = -14

x + 5 = 9

x = 4

Best of Luck!

5 0
3 years ago
Read 2 more answers
Violet puts 12 out of every 192 dollars that she earns into a savings account. Which of the following is equivalent to
Paladinen [302]

Answer:

I think it is 12

Step-by-step explanation:

6 0
3 years ago
Why is tangent's period 180 degrees?
Ostrovityanka [42]
In order for tan x = 0, sin x must be = 0 
<span>sin x = 0 at 0 degrees and 180 degrees </span>
<span>so, tan x = 0 at 0, pi </span>
<span>as for the period being pi, you need to have tan x be negative and positive </span>
<span>so remember the acronym All students take calculus ( ASTC) for where they are positive. </span>
<span>tan x is negative in Quadrant II and positive in Quadrant I, since it just repeats after that, the period therefore must be Pi.</span>
3 0
3 years ago
A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one
TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are \binom44\binom{48}1=48 possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by \binom{13}1=13. So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing \binom41\binom{48}4=778,320 possible hands. Exactly 2 aces are drawn in \binom42\binom{48}3=103,776 hands. And so on. This gives a total of

\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656

possible hands containing at least 1 ace, and hence B occurs with probability

\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. P(A\cap B)=P(A)P(B). This happens if

  • the hand has 4 aces and 1 non-ace, or
  • the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So A\cap B consists of 96 possible hands, which occurs with probability

\dfrac{96}{\binom{52}5}\approx0.0000369

and so the events A and B are NOT independent.

4 0
3 years ago
On a coordinate plane, a rectangle has points (3, negative 1), (3, negative 5), (negative 6, negative 5), (negative 6, negative
Natalka [10]
The answer is 9 times 4

8 0
3 years ago
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