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Elan Coil [88]
1 year ago
12

Determine the multiplicity of the roots of the function k(x) = x(x + 2)3(x + 4)2(x − 5)4. 0, -2, -4, 5

Mathematics
2 answers:
mote1985 [20]1 year ago
5 0

Answer:

0 has multiplicity 1

−2 has multiplicity 3

−4 has multiplicity 2

5 has multiplicity 4

Sergeu [11.5K]1 year ago
4 0

Answer:

The polynomial function k(x)=x(x+2)^3(x+4)^2(x-5)^4

To determine the multiplicity of 0, -2, -4, 5.

The multiplicity of a root is the number of times the root appears.

First find the root of the equation, set the function equals to zero.

x(x+2)^3(x+4)^2(x-5)^4=0

therefore, the root of this function are, x=0,-2, -4, 5

To find the multiplicity of the roots:

A factor of x would have a root at x=0 with multiplicity of 1

similarly,  x=-2 with multiplicity of 3

x=-4 with multiplicity of 2

x=5 with multiplicity of 4.



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First let's write out the inequality before choosing a graph.

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For questions 2-5, the number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 a
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The number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 and a standard deviation of 2.12.

a)What is the z-score value of a randomly selected bag of Skittles that has 35 Skittles? a) 1.62 b) -1.62 c) 3.40 d) -3.40 e)1.303.

The formula for calculating a z-score is is z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

z = 35 - 38.4/2.12

= -1.60377

Option b) -1.62 is correct

b) What is the probability that a randomly selected bag of Skittles has at least 37 Skittles? a) .152 b) .247 c) .253 d).747e).7534. .

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Mean of 38.4 and a standard deviation of 2.12.

z = (37 - 38.4)/2.12

= -0.66038

P-value from Z-Table:

P(x<37) = 0.25451

The probability that a randomly selected bag of Skittles has at least 37 Skittles is 0.25451

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c) What is the probability that a randomly selected bag of Skittles has between 39 and 42 Skittles? a) .112 b) .232 c) .344 d).457 e).6125.

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

For 39 Skittles

z = (39 - 38.4)/2.12

= 0.28302

Probability value from Z-Table:

P(x = 39) = 0.61142

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z = (42 - 38.4)/2.12

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Probability value from Z-Table:

P(x = 42) = 0.95526

The probability that a randomly selected bag of Skittles has between 39 and 42 Skittles is:

P(x = 42) - P(x = 39

0.95526 - 0.61142

0.34384

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d) What is the percentile rank of a randomly selected bag of Skittles that has 40 Skittles in it? a)82nd b) 78th c) 75th d)25th e)22nd

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (40 - 38.4)/2.12

= 0.75472

P-value from Z-Table:

P(x = 40) = 0.77479

Converting to percentage = 0.77479× 100

= 77. 479%

≈ 77.5

Percentile rank = 78th

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