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3 years ago

Can you help me please? thank you <3

Mathematics

1 answer:

olganol [36]3 years ago

7 0

Welcome to the era of LeBron. If you thought he was doing the game a disservice by being a sore loser and egomaniac, you haven’t seen anything yet. It wasn’t enough that he decided to become a pariah, he and his group of cronies have decided they want to ruin the NBA.

How? By systematically turning every superstar in the NBA into LeBron James. They are trying to turn them into spoiled, delusional, lying, quitting cowards.

Two of the biggest stories this offseason have been the drama in Denver and New Orleans. Both players, for whatever reasons, have decided that they no longer want to be the cornerstone of a franchise. Before the summer of Lebron, neither had really been asking to get out of their situations. So what is the common thread here? Well, both are represented by agent Leon Rose. Who else is represented by Rose? Why Mr. James, of course.

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In the rectangle below, AE = 3x+7, CE = 5x-7, and m Find BE and m

Alborosie

Answer:

AE-3*x+7=0 and CE=5*x-7

Step-by-step explanation:

-3*x and -7 move these terms to the left it becomes AE-3*x-7=0 do the same step that you did with the first one.

3 0

3 years ago

Find the h.c.f of {x}^{3} + {y}^{3} \: and \: \: {x}^{2} - xy + {y}^{2}x 3 +y 3 andx 2 −xy+y 2

Ahat [919]

Factorize the sum of cubes:

$x^3 + y^3 = (x+y) (x^2 - xy + y^2)$

so $x^2-xy+y^2$ is a factor of $x^3+y^3$ . Then the HCF is $\boxed{x^2 - xy + y^2}$ .

5 0

2 years ago

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:

<em>(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?</em>

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.

5 0

3 years ago

Help me with this please

nataly862011 [7]

Answer: Spanish

Step-by-step explanation:

Spanish

7 0

3 years ago

Read 2 more answers

Which is the better buy?

madam [21]

2-gallon
bucket for 66.72

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3 years ago