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s2008m [1.1K]
3 years ago
12

Distance between riverside and milton if they are 12 cm apart on a map with a scale of 4cm : 21 (?

Mathematics
1 answer:
solniwko [45]3 years ago
7 0
You dont have anything that we can answer to.
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In how many ways can we seat 3 pairs of siblings in a row of 7 chairs, so that nobody sits next to their sibling
monitta

Answer:

1,968

Step-by-step explanation:

Let x₁ and x₂, y₁ and y₂, and z₁ and z₂ represent the 3 pairs of siblings, and let;

Set X represent the set where the siblings x₁ and x₂ sit together

Set Y represent the set where the siblings y₁ and y₂ sit together

Set Z represent the set where the siblings z₁ and z₂ sit together

We have;

Where the three siblings don't sit together given as X^c∩Y^c∩Z^c

By set theory, we have;

\left | X^c \cap Y^c \cap Z^c  \right | = \left | X^c \cup Y^c \cup Z^c  \right | =  \left | U  \right | - \left | X \cup Y \cup Z  \right |

\left | U  \right | - \left | X \cup Y \cup Z  \right | = \left | U  \right | - \left (\left | X \right | +  \left | Y\right | +  \left | Z\right | -  \left | X \cap Y\right | -  \left | X \cap Z\right | -  \left | Y\cap Z\right | +  \left | X \cap Y \cap Z\right | \right)

Therefore;

\left | X^c \cap Y^c \cap Z^c  \right | = \left | U  \right | - \left (\left | X \right | +  \left | Y\right | +  \left | Z\right | -  \left | X \cap Y\right | -  \left | X \cap Z\right | -  \left | Y\cap Z\right | +  \left | X \cap Y \cap Z\right | \right)

Where;

\left | U\right | = The number of ways the 3 pairs of siblings can sit on the 7 chairs = 7!

\left | X\right | = The number of ways x₁ and x₂ can sit together on the 7 chairs = 2 × 6!

\left | Y\right | = The number of ways y₁ and y₂ can sit together on the 7 chairs = 2 × 6!

\left | Z\right | = The number of ways z₁ and z₂ can sit together on the 7 chairs = 2 × 6!

\left | X \cap Y\right | = The number of ways x₁ and x₂ and y₁ and y₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | X \cap Z\right | = The number of ways x₁ and x₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | Y \cap Z\right | = The number of ways y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | X \cap Y \cap Z\right | = The number of ways x₁ and x₂,  y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 2 × 4!

Therefore, we get;

\left | X^c \cap Y^c \cap Z^c  \right | = 7! - (2×6! + 2×6! + 2×6! - 2 × 2 × 5! - 2 × 2 × 5! - 2 × 2 × 5! + 2 × 2 × 2 × 4!)

\left | X^c \cap Y^c \cap Z^c  \right | = 5,040 - 3072 = 1,968

The number of ways where the three siblings don't sit together given as \left | X^c \cap Y^c \cap Z^c  \right |  = 1,968

5 0
3 years ago
How many carbon monoxide molecules are in 0.75 moles of carbon monoxide?
Mrac [35]

No of molecules = no of moles x avogadro’s number

0.75 ( 6.02 × 10²³) = 4.515 x 10²³ (standard from)

7 0
3 years ago
Abbie decides to start doing crunches as part of her daily workout. She decides to do 15 crunches the first day and then increas
Pavlova-9 [17]
For this case, the first thing we must do is define a variable.
 We have then:
 n: number of days.
 We now write the explicit formula that represents the problem.
 We have then:
 an = 4n + 15
 Where,
 15: crunches the first day
 4: increase the number 4 each day
 Answer:
 
An explicit formula for the number of crunches Abbie will do on day n is:
 
an = 4n + 15
3 0
3 years ago
Sidney is making greeting cards, which she will sell by the box at an arts fair. She paid $32 for a booth at the fair, and the m
borishaifa [10]

So think of it like this she starts off $32 negative and subtract 3 from each box of cards she makes . So each box of cards she will make $8 that being said

32/8 =4 (it will take 4 box of cards to pay off her booth expense).

Sales would be 11x 4 =44

Expenditures would be 3x4 =32 (the amount used to make cards ) +32(the amount paid for the booth)

Which in total would be $64

4 0
3 years ago
25.7% as a fraction<br> plz help
vlada-n [284]

Answer:

257/1000

Step-by-step explanation:

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

4 0
3 years ago
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