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jolli1 [7]
2 years ago
5

Lena drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 6 hours. When Lena drove h

ome, there was no traffic and the trip only took 4 hours. If her average rate was 22 miles per hour faster on the trip home, how far away does Lena live from the mountains?
Mathematics
1 answer:
tekilochka [14]2 years ago
7 0
I believe the answer is 88
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Answer quick plsssssssssss
Deffense [45]

Answer:

$64

Step-by-step explanation:

Let the pay of Bella be x and Glen be y.

ATQ, 7x-5y=46 and 6x+5y=188, x=18, y=16.

Glen earned 4*16=64 bucks on Friday.

5 0
2 years ago
If a graph passes the horizontal line test, but not the vertical line test, is it still a function?
nydimaria [60]

If a graph fails the vertical line test it's not a function; you can't have a single x map to two ys.

Vice versa, if the graph passes the verticle line test, it's a function.  

The horizontal line test tests for injectivity aka one-to-one-ness; it has no bearing on whether something's a function.

8 0
2 years ago
A small theater had 9 rows of 25 chairs each. An extra 7 chairs have just been brought in. How many chairs are in the
Alchen [17]

Answer:

232

Step-by-step explanation:

9 x 25 = 225   225 + 7 = 232

another way to find it

9 x 20 = 180   9 x 5 = 45   180 + 45 = 225 + 7 = 232

8 0
2 years ago
Simplify(4-x) (7+x) <br>​
Sophie [7]

(4-x) ( 7+ ×)

28 +4x -7x -x²

28 -3x - x²

7 0
2 years ago
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
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