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ra1l [238]
3 years ago
6

Use synthetic substitution to find g(2) and g(-7) for the function g(x)=5x^4 - 3x^2 +6x-4

Mathematics
1 answer:
Nina [5.8K]3 years ago
8 0
G(x) = 5x^4 - 3x^2 + 6x - 4 

g(2) = 5(2^4) - 3(2^2) + 6(2) - 4
g(2) = 5(16) - 3(4) + 12 - 4
g(2) = 80 - 12 + 12 - 4
g(2) = 76

g(-7) = 5(-7^4) - 3(-7^2) + 6(-7) - 4
g(-7) = 5(2401) - 3(49) - 42 - 4
g(-7) = 12005 - 147 - 42 - 4
g(-7) = 12005 - 193
g(-7) = 11812
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7) In Problem 6 we poured a large and a small bowl of cereal from a box. Suppose the amount of cereal that the manufacturer puts
olga_2 [115]

Question:

The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and pour one large and one small bowl.

Answer:

a) The expected amount of cereal left in the box is 12.2 ounces

b) The standard deviation  \sigma_{x+y+z}, is 0.5099

c) In a Normal model, the probability that the box still contains more than 13 ounces is P(Z-(X+Y) > 13) = 5.821 %.

Step-by-step explanation:

Let X represent the amount of cereal that can be poured into a small bowl and Y represent the amount of cereal that can be poured into a large bowl and Z represent the amount of cereal that the manufacturer puts in the box, then the expected amount of cereal left in the box is given by

Z - (X + Y)

(a) The expected amount of cereal left in the box is given as

P(Z - (X + Y)) = μ = μ_Z - μ_X - μ_Y = 16.2 - 1.5 - 2.5 = 12.2 ounces

The expected amount of cereal left in the box = 12.2 ounces

b) The standard deviation is given by  the root of the sum of the variance

That is

\sigma_{x+y+z} ^2 = \sigma_x^2 + \sigma_y^2 +\sigma_z^2 and

\sqrt{\sigma_{x+y+z} ^2}  = \sqrt{\sigma_x^2 + \sigma_y^2 +\sigma_z^2} =\sqrt{0.1^2+0.4^2+0.3^2} = 0.5099

The standard deviation,  \sigma_{x+y+z}, = 0.5099

c) The probability that the box still contains more than 13 ounces is given by

P(Z-(X+Y) > 13)

Where z-score is  

z=\frac{x-\mu}{\sigma} = \frac{13-12.2}{0.5099}=  1.5689 ≈ 1.57

From the z-score table P(Z = 1.57) = 0.94179

Therefore the probability of the box containing ≤ 13 is 0.94179, that is

P(Z-(X+Y) ≤ 13) = 0.94179 and

P(Z-(X+Y) > 13) = 1 - 0.94179 = 0.05821 = 5.821 %.

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