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hodyreva [135]
3 years ago
8

ASAP What is the multiplicity of the root (x^3-1)?

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
5 0
Answer: =(x - 1) x (x^2 + x + 1)
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aleksklad [387]
7,000 ones is the answer
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(i) 809, 708, 607, _____, _ class 4<br>__class 4​
schepotkina [342]

Answer:

809, 708, 607, 506, 405, 304, 203, 102, 1, -100

Step-by-step explanation:

809, 708, 607, _____, _______

First term = 809

Second term = 708

Third term = 607

Difference between first term and second term = 809 - 708

= 101

Difference between second term and third term = 708 - 607

= 101

Therefore, the common difference is 101

Fourth term = 607 - 101

= 506

Fifth term = 506 - 101

= 405

Sixth term = 405 - 101

= 304

Seventh term = 304 - 101

= 203

Eighth term = 203 - 101

= 102

Ninth term = 102 - 101

= 1

Tenth term = 1 - 101

= - 100

809, 708, 607, 506, 405, 304, 203, 102, 1, -100

4 0
3 years ago
Help me PLEASE I need HELP
aivan3 [116]

Answer:

x=20

Step-by-step explanation:

180=2x+6x+20

180=8x+20

160=8x

20=x

5 0
3 years ago
What’s the scientific notation for .039
trasher [3.6K]

Answer: 3.9 x 10-2

Step-by-step explanation:

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4 years ago
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A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air
Stells [14]

Answer:

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

H₀ : <em>p = </em>0.6

H₁ : <em>p  = </em>0.6, this explains the acceptance region as;

p° ≤ \frac{315}{500}=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

   = P (p°>0.63 | p=0.6)

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

   

    = P  [\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]

    = P  [\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]

    = P   [Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]

    = P   [Z > 1.37]

    = 1 - P   [Z ≤ 1.37]

    = 1 - Ф (1.37)

    = 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

    ≅ 0.0853

b)

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

  = P [p° ≤ 0.63 | p = 0.75]

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

  = P [\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]

  = P [\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P [Z ≤ -6.20]

  = Ф (-6.20)

  ≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

6 0
3 years ago
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