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Aleks04 [339]
2 years ago
6

A basketball player makes 70% of the free throws he shoots. Suppose that he tries 15 free throws.

Mathematics
1 answer:
Volgvan2 years ago
8 0

Answer:

a.) .95

b.) The expected number of baskets is 10.50.

c.) 1.7748

Step-by-step explanation:

a.) This is a binomial distribution as there are two possibilities: makes a free throw or doesn't. This means that you can use the binomial function on a calculator to figure out the answer. Use the binomial CDF function on a calculator and the number of trials=15, probability of success=.7, lower bound=0, upper bound=7. Once you have evaluated the answer, .0500, it will need to be subtracted from1, as you want everything not included in this section. The answer to part a is thus 1-.0500=.95.

b.) The expected value is calculated by taking the total number of shots and multiplying it by the probability of making the shot: 15×.7=10.5 shots.

c.) The standard deviation of a binomial distribution can be calculated by the formula \sqrt{(sample size)(probability)(1-probability)}. Plugging in the numbers you get \sqrt{(15)(.7)(.3)}=1.7748.

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3 0
3 years ago
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Step-by-step explanation:

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3 0
3 years ago
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Answer:

First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

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2

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9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

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1

+

R

2

=

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2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

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−

18

9

⎤

⎥

⎦

−

3

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1

+

R

3

=

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3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

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R

2

and

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3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

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15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

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3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

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x

−

y

+

z

=

8

y

−

12

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=

−

15

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=

1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

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x

−

y

+

z

=

8

y

−

12

z

=

−

15

z=1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

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