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Given f(x) = 5x4 – x2 + 6x – 1. What is Limit of f (x) as x approaches negative 1?

irinina [24]

Answer:

-3

Step-by-step explanation:

5 0

3 years ago

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A basketball player scored x, y, and z points in 3 games. Express the average number of points scored by the player in terms of

Anettt [7]

The average will be the total points divided by the number of games:

(x+y+z) / 3

7 0

4 years ago

The spoke of a wheel reaches from the center of the wheel to its rim. If the circumference of the rim of

never [62]

C = 2(pi)r....so we are looking for the radius (r)
42 = 2 * 3.14 * r
42 = 6.28r
42/6.28 = r
6.69 = r <===

4 0

3 years ago

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Find the area of the shaded region? 3x + 1 3x + 1

mash [69]

Im pretty sure its 3x + 1

7 0

3 years ago

7.2 Given a test that is normally distributed with a mean of 100 and a standard deviation of 10, find: (a) the probability that

kompoz [17]

Answer:

a)

The probability that a single score drawn at random will be greater than 110

P( X > 110) = 0.1587

b)

The probability that a sample of 25 scores will have a mean greater than 105

P( x> 105) = 0.0062

c)

The probability that a sample of 64 scores will have a mean greater than 105

P( x⁻> 105) = 0.002

d)

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = 0.9544

Step-by-step explanation:

a)

Given mean of the Normal distribution 'μ' = 100

Given standard deviation of the Normal distribution 'σ' = 10

a)

Let 'X' be the random variable of the Normal distribution

let 'X' = 110

$Z = \frac{x-mean}{S.D} = \frac{110-100}{10} =1$

The probability that a single score drawn at random will be greater than 110

P( X > 110) = P( Z >1)

= 1 - P( Z < 1)

= 1 - ( 0.5 +A(1))

= 0.5 - A(1)

= 0.5 -0.3413

= 0.1587

b)

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{25} } } = 2.5$

The probability that a single score drawn at random will be greater than 110

P( x> 105) = P( z > 2.5)

= 1 - P( Z< 2.5)

= 1 - ( 0.5 + A( 2.5))

= 0.5 - A ( 2.5)

= 0.5 - 0.4938

= 0.0062

The probability that a single score drawn at random will be greater than 105

P( x> 105) = 0.0062

c)

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{64} } } = 4$

The probability that a single score drawn at random will have a mean greater than 105

P( x> 105) = P( z > 4)

= 1 - P( Z< 4)

= 1 - ( 0.5 + A( 4))

= 0.5 - A ( 4)

= 0.5 - 0.498

= 0.002

The probability that a sample of 64 scores will have a mean greater than 105

P( x⁻> 105) = 0.002

d)

Let x₁ = 95

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{95-100}{\frac{10}{\sqrt{16} } } = -2$

Let x₂ = 105

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{16} } } = 2$

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = P( -2≤z≤2)

= P(z≤2) - P(z≤-2)

= 0.5 + A( 2) - ( 0.5 - A( -2))

= A( 2) + A(-2) (∵A(-2) =A(2)

= A( 2) + A(2)

= 2 × A(2)

= 2×0.4772

= 0.9544

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = 0.9544

7 0

3 years ago