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3 years ago

What separates two sides of an equation

Mathematics

1 answer:

Studentka2010 [4]3 years ago

4 0

Answer:

the equal sign

Step-by-step explanation:

expression = answer

this splits it

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Instructions for a craft project say that the length of a piece of red ribbon should be 7 inches less than the length of a piece

KIM [24]

Answer:

12 inches of red ribbon + 7 inches

therefore the length of a blue ribbon is = 19 inches

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3 years ago

Andrew believes the honor roll students at his school have an unfair advantage in being assigned to the math class they request.

Artist 52 [7]

Honor roll Not on honor roll Total
Received math class requested 315 64 379
Did not get math class requested 41 80 121
Total 356 144 500

Honor roll: request granted: 315/356 = 0.88 x 100% = 88%
Not Honor roll request granted: 64/144 = 0.44 x 100% = 44%

Honor roll students were given preference in granting request than those not in the honor roll.

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3 years ago

Ryan is planning a party for his parents' anniversary. The caterer charges a delivery fee of $14 plus an additional & 5.25 p

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Answer:

Step-by-step explanation:

$14+$5.25=$19.25

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3 years ago

Read 2 more answers

Solve each equation r - 5 1/6 = 10 please help me Thank youuu!

Charra [1.4K]

[tex}r - 5 \frac{5}6 + 5\frac{5}6 = 10 +5\frac{5}6{/tex] $r=15 \frac{5}6$

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3 years ago

Read 2 more answers

Exercise 12.1.2: The probability of an event under the uniform distribution - random permutations. About A class with n kids lin

Nataly [62]

Answer:

a) $P_a=\frac{1}{n}$

b) $P_b=\frac{1}{n(1-n)}$

c) $P_c=\frac{2}{n}$

Step-by-step explanation:

The question is incomplete:

(a) What is the probability that Celia is first in line? (b) What is the probability that Celia is first in line and Felicity is last in line? (c) What is the probability that Celia and Felicity are next to each other in the line?

a) The probability that Celia is first in line, if there are n kids and all of them have the same chance, is 1/n.

$P_a=\frac{1}{n}$

b) The probability that Celia is first in line and Felicity is last in line is

$P_b=\frac{1}{n} \frac{1}{n-1} =\frac{1}{n(1-n)}$ .

It is deducted like that:

If Celia is placed in the first line (what has a probablity of 1/n), there are left (n-1) kids. Then, the probability of placing Felicity in the last place is 1/(n-1).

Both probabilities multiplied give 1/(n*(n-1)).

c) The probability that Celia and Felicity are next to each other in the line is

$P_c=\frac{2(n-1)!}{n!}=\frac{2}{n}$

There are n! combinations for kids lines, where (n-1)! permutations have Celia before Felicity and other (n-1)! permutations have Felicity before Celia.

Then, we have 2(n-1)! permutations that have Celia and Felicity next to each other, over n! permutations possible.

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3 years ago