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padilas [110]
3 years ago
15

A basketball player averages 13.5 points per game. Use this rate to find the value of x, the number of points scored in 7 games.

Mathematics
2 answers:
pychu [463]3 years ago
6 0

Answer:

Knowing that a ratio applies to the same situation in just a larger/smaller scale; 

(13.5)(7)=x (where x is the predicted number of scores) 

94.5=x 

Therefore, the player should score 94.5 points in 7 games. 

Hope I helped :) 

Read more on Brainly.com - brainly.com/question/10662889#readmore

Step-by-step explanation:

xxTIMURxx [149]3 years ago
4 0

Knowing that a ratio applies to the same situation in just a larger/smaller scale; 


(13.5)(7)=x (where x is the predicted number of scores) 

94.5=x 


Therefore, the player should score 94.5 points in 7 games. 


Hope I helped :) 

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In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A  

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

4 0
3 years ago
Mrs. Jackson gives the table below to her students.
krek1111 [17]
M= 20 because the rate of change is -3.
5 0
3 years ago
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Which is the best estimate of x based on rounding the constants and coefficients in the equation to the nearest integer 6.15 (x
artcher [175]
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6.15x = 76.32
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x = 12.41
7 0
3 years ago
The circumference of a circle is 704 cm,find its area
Digiron [165]
The circumferince of the circle is 2πR ;
The area of the circle is πR^{2};
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5 0
3 years ago
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Solve 8x+2y=13 and 4x+y=11 using substitution
Aleonysh [2.5K]

Answer:

No Solution

Step-by-step explanation:

8x+2y=13                         4x+y=11               Subtract 4x on both sides to isolate y

                                        y=-4x+11       <---- Substitute this equation into the                                            

                                                                     equation 8x+2y=13

8x+2(-4x+11)=13

8x-8x+22=13

22=13 (not true)

No Solution          

                     

4 0
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