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krek1111 [17]
3 years ago
11

Exhibit 6-2 the weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 poun

ds. 26. refer to exhibit 6-2. what is the minimum weight of the middle 95% of the players?
a. 196
b. 151
c. 249
Mathematics
1 answer:
Evgen [1.6K]3 years ago
5 0
The question is asking for the lower bound of the 95% two tailed Confidence interval of the normally distributed population.

95% C.I. is given by 200 + or - 1.96(25) = 200 + or - 49 = (151, 249)
Therefore, the minimum weight of the middle 95% of players is 151 pounds.
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garri49 [273]

Answer:

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Step-by-step explanation:

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faltersainse [42]

The price of one vegetarian special lunch is $7 and price of one chicken special lunch is $8.

Step-by-step explanation:

Let,

Price of one vegetarian special lunch = x

Price of one chicken special lunch = y

According to given statement;

21x+40y=467      Eqn 1

28x+36y=484     Eqn 2

Multiplying Eqn 1 by 28

28(21x+40y=467)\\588x+1120y=13076\ \ \ Eqn\ 3

Multiplying Eqn 2 by 21

21(28x+36y=484)\\588x+756y=10164\ \ \ Eqn\ 4

Subtracting Eqn 4 from Eqn 3

(588x+1120y)-(588x+756y)=13076-10164\\588x+1120y-588x-756y=2912\\364y=2912

Dividing both sides by 364

\frac{364y}{364}=\frac{2912}{364}\\y=8

Putting y=8 in Eqn 1

21x+40(8)=467\\21x+320=467\\21x=467-320\\21x=147

Dividing both sides by 21

\frac{21x}{21}=\frac{147}{21}\\x=7

The price of one vegetarian special lunch is $7 and price of one chicken special lunch is $8.

Keywords: linear equation, elimination method

Learn more about elimination method at:

  • brainly.com/question/10081622
  • brainly.com/question/10341324

#LearnwithBrainly

5 0
3 years ago
ILL MARK BRAINLY!!
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Answer:

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Step-by-step explanation:

When you put something in parenthesis by a number, your multiplying each individual number by the number outside of parenthesis. Therefore, 5(2) = 10 and 5(7) = 35 and that equals 45, and 5(2+7) you distribute the 5 to both numbers and get 10+35 which equals 45 aswell.

7 0
3 years ago
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Find the excluding values for the equations a^3+3a^2-10a/5a^3-20a
anzhelika [568]

Answer:

a+5/5(a+2)

Step-by-step explanation:

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7 0
3 years ago
EXAMPLE 1 (a) Find the derivative of r(t) = (2 + t3)i + te−tj + sin(6t)k. (b) Find the unit tangent vector at the point t = 0. S
Tatiana [17]

The correct question is:

(a) Find the derivative of r(t) = (2 + t³)i + te^(−t)j + sin(6t)k.

(b) Find the unit tangent vector at the point t = 0.

Answer:

The derivative of r(t) is 3t²i + (1 - t)e^(-t)j + 6cos(6t)k

(b) The unit tangent vector is (j/2 + 3k)

Step-by-step explanation:

Given

r(t) = (2 + t³)i + te^(−t)j + sin(6t)k.

(a) To find the derivative of r(t), we differentiate r(t) with respect to t.

So, the derivative

r'(t) = 3t²i +[e^(-t) - te^(-t)]j + 6cos(6t)k

= 3t²i + (1 - t)e^(-t)j + 6cos(6t)k

(b) The unit tangent vector is obtained using the formula r'(0)/|r(0)|. r(0) is the value of r'(t) at t = 0, and |r(0)| is the modulus of r(0).

Now,

r'(0) = 3t²i + (1 - t)e^(-t)j + 6cos(6t)k; at t = 0

= 3(0)²i + (1 - 0)e^(0)j + 6cos(0)k

= j + 6k (Because cos(0) = 1)

r'(0) = j + 6k

r(0) = (2 + t³)i + te^(−t)j + sin(6t)k; at t = 0

= (2 + 0³)i + (0)e^(0)j + sin(0)k

= 2i (Because sin(0) = 0)

r(0) = 2i

Note: Suppose A = xi +yj +zk

|A| = √(x² + y² + z²).

So |r(0)| = √(2²) = 2

And finally, we can obtain the unit tangent vector

r'(0)/|r(0)| = (j + 6k)/2

= j/2 + 3k

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3 years ago
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