3x+2y I don't get need help

Factorise quadratic equation<br> n^2-n=0

Sindrei [870]

n² - n = 0

n(n-1)

$\\ \\$

7 0

3 years ago

PLZZZZZZ IL do anything give brainilest 5 star i'll like My grade depends on this PLEASE I NEED THIS TO BE CORRECT AND SHOW YOUR

sesenic [268]

Answer:

$195.00

Step-by-step explanation:

perimeter is 54 feet

area is 152 $ft^{2}$

54' * $1.50 = $81.00

152/8 = 19 bags of mulch needed

19*$6.00= $114.00

$195.00 you get away from the store for under 200 :P

P.S. since you said you'll do anything, send me $1,000,000.00 dollars :D

8 0

3 years ago

Given g(x)=8x+3 find the following g(-1) and g(2a+1)

Degger [83]

Answer:

g(-1) = -5

g(2a+1) = 16a+11

Step-by-step explanation:

Substitute the given value into the function and evaluate.

So for g(x) = 8x+3

You substitute g(-1)

So g(-1) = 8(-1)+3

That is how you get -5

Then g(2a+1)

So g(2a+1) = 8(2a+1)+3

Thats how you get 16a+11

Hope this helps. Mark as brainlist

4 0

3 years ago

Helpppppp I’ll Mark you brainliest

Debora [2.8K]

Answer:

6.32

Step-by-step explanation:

:|

3 0

3 years ago

Read 2 more answers

PLEASE HELP! The table shows the number of championships won by the baseball and softball leagues of three youth baseball divisi

Irina18 [472]

Answer:

Question 1: $P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16}}{ \frac{4}{16}} = \frac{1}{2}$

Question 2:

A. $P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

B. $P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

C. $P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

Step-by-step explanation:

Conditional probability is defined by

$P(A|B)= \frac{P(A and B)}{P(B)}$

with P(A and B) beeing the probability of both events occurring simultaneously.

Question 1:

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16} }{ \frac{4}{16} } = \frac{1}{2}$

Question 2.A:

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

Question 2.B:

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( Z and B)= \frac{ 1 }{ 16 }[/tex]

By definition,

$P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

Question 3.B

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

then

$P (Y or Z) = P(Y) + P(Z) = \frac{6}{16}$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

so

$P((YorZ) and B)= \frac{3}{16}$

By definition,

$P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

3 0

3 years ago