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nikdorinn [45]
3 years ago
5

A set of final examination grades in an introductory statistics course is normally​ distributed, with a mean of 77 and a standar

d deviation of 7. Complete parts​ (a) through​ (d). a. What is the probability that a student scored below 86 on this​ exam? The probability that a student scored below 86 is nothing. ​(Round to four decimal places as​ needed.)
Mathematics
1 answer:
Shalnov [3]3 years ago
7 0

Answer:

0.9007 is the probability that a student scored below 86 on this​ exam.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 77

Standard Deviation, σ = 7

We are given that the distribution of examination grades is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(student scored below 86)

P(x < 86)

P( x < 86) = P( z < \displaystyle\frac{86 - 77}{7}) = P(z < 1.2857)

Calculation the value from standard normal z table, we have,  

P(x < 1.2857) = 0.9007 = 90.07\%

0.9007 is the probability that a student scored below 86 on this​ exam.

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<h3>Answer:</h3>

C) y = 6x

<h3>Step-by-step explanation:</h3>

Pick any point. It is often convenient to use x = 1 (no marked point) or x = 10 (where y = 60).

Use these values to see which equation agrees.

A: 60 ≠ (1/6)·10

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D: 60 ≠ 12·10

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Or, you can solve ...

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How to find the circumference of a circle​
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let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep
Nutka1998 [239]

Answer with Step-by-step explanation:

We are given that

x_1,x_2 and x_3 are linearly independent.

By definition of linear independent there exits three scalar a_1,a_2 and a_3 such that

a_1x_1+a_2x_2+a_3x_3=0

Where a_1=a_2=a_3=0

y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1

We have to prove that y_1,y_2 and y_3 are linearly independent.

Let b_1,b_2 and b_3 such that

b_1y_1+b_2y_2+b_3y_3=0

b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0

b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0

(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0

b_1+b_3=0

b_1=-b_3...(1)

b_1+b_2=0

b_1=-b_2..(2)

b_2+b_3=0

b_2=-b_3..(3)

Because x_1,x_2 and x_3 are linearly independent.

From equation (1) and (3)

b_1=b_2...(4)

Adding equation (2) and (4)

2b_1==0

b_1=0

From equation (1) and (2)

b_3=0,b_2=0,b_3=0

Hence, y_1,y_2 and y_3 area linearly independent.

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Natasha2012 [34]

Answer:

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Answer:

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Step-by-step explanation:

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So, the difference is (56.87 - 55.98) = 0.89 seconds.

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5 0
3 years ago
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