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DanielleElmas [232]
2 years ago
15

A sports statistician was interested in the relationship between game attendance (in thousands) and the number of

Mathematics
1 answer:
Luda [366]2 years ago
4 0

Answer:

The predicted number of wins for a team that has an attendance of 2,100 is 25.49.

Step-by-step explanation:

The regression  equation for the relationship between game attendance (in thousands) and the number of  wins for baseball teams is as follows:

\hat y = 4.9\cdot x + 15.2

Here,

<em>y</em> = number of wins

<em>x</em> = attendance (in thousands)

Compute the number of wins for a team that has an attendance of 2,100 as follows:

\hat y = 4.9\cdot x + 15.2

  =(4.9\times 2.1)+15.2\\=10.29+15.2\\=25.49

Thus, the predicted number of wins for a team that has an attendance of 2,100 is 25.49.

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Step-by-step explanation:

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Together Mary, Jane, and Ruth received $30. Mary received $6, Jane received $9, and Ruth received $15. What percent of the $30 d
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Hey there! :D 

So, we know that Jane received $9 out of the $30. 

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9/30= .3

Turn that into a percentage. 

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We can check this by finding the other percentages. 

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~kaikers 
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2 years ago
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Find the GCF of 44j5k4 and 121j2k6.
BartSMP [9]

Answer:

D. 11j^2k^4

Step-by-step explanation:

We are asked to find the GCF of 44j^5k^4\text{ and }121j^2k^6.

Since we know that GCF of two numbers is the greatest number that is a factor of both of them.

First of all we will GCF of 44 and 121.

Factors of 44 are: 1, 2, 4, 11, 22, 44.

Factors of 121 are: 1, 11, 11, 121.

We can see that greatest common factor of 44 and 121 is 11.

Now let us find GCF of j^5\text{ and }j^2.

Factors of j^5 are: j*j*j*j*j

Factors of j^2 are: j*j

We can see that greatest common factor of j^5\text{ and }j^2 is j*j=j^2.

Now let us find GCF of k^4\text{ and }k^6.

Factors of k^4 are: k*k*k*k    

Factors of k^6 are:k*k*k*k*k*k

We can see that greatest common factor of  k^4\text{ and }k^6 is k*k*k*k=k^4.

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Answer:

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Step-by-step explanation:

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