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ira [324]
3 years ago
6

Sergei buys a rectangular rug for his living room. He measures the diagonal of the rug to be 18 feet. The length of the rug is 3

feet longer than the width. What are the approximate dimensions of the rug? Round each dimension to the nearest tenth of a foot.
A) 10.4 feet by 7.4 feet
B) 11.1 feet by 8.1 feet
C) 13.4 feet by 10.4 feet
D) 14.1 feet by 11.1 feet
Mathematics
1 answer:
kvv77 [185]3 years ago
5 0
The correct answer is d. Please give me brainlest I hope this helps let me know if it’s correct or not okay
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inna [77]

Answer:

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56738w7421098548091672949123749279102

Step-by-step explanation:

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3 years ago
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Degger [83]

Answer:

an=7n-19

(the last one)

Step-by-step explanation:

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3 years ago
Anya has a pink ribbon and a purple ribbon. The pink ribbon is 6 ⅕ inches long. The purple ribbon is 8 ²⁄₁₀ inches. How much rib
balu736 [363]

Answer:

B. 14 ⁴⁄₁₀

Step-by-step explanation:

I say its b because all u have to do it multiply  6 ⅕  but not the 6, just the 1 and the 5, just multiply by two and both fractions will have the same total amount which is 10 and 6 ⅕ will turn to 6 with 2/10 so all u have to do now is add that fraction to 8 ²⁄₁₀

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3 years ago
S = the team's score
Akimi4 [234]

Answer:

s would be the dependent variable, and g the independent.

Step-by-step explanation:

Whatever the score is, it would depend on how many goals the team has scored.

6 0
3 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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