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natima [27]
3 years ago
15

Please help [99 points]

Mathematics
2 answers:
dimaraw [331]3 years ago
8 0

Hello!

I believe the answer is C) Nonlinear function because although it passes through the origin, it isn't a straight line. Therefor it is a Nonlinear function.

I hope it helps!

Luda [366]3 years ago
5 0
The graph shows a nonlinear function. This is because the line is not straight. I hope this helps :) <span />
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Sliva [168]

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C

Step-by-step explanation:

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What is the vertical distance from the origin to the point (2, 9)?
vladimir2022 [97]

Answer:

The vertical distance from the origin to the point (2,9) is 9

Or

9.2195

Step-by-step explanation:

It's 9 because y coordinate gives the vertical distance.

Or

It might be 9.2195 because they tell us the point is 2.9 and they are asking the distance from the origin which is zero. So we can use Pythagorean theorem. A²+B²=C² Look at it as a triangle. They have given us A and B, so you can find C² using those information. If you plot the point in the line you know that A=2 and B=9. Using the formula will be 2²+9²=C² ⇒ 4+81=C² ⇒ 85=C². To get rid of the power of 2 in c we have to square both side to cancel it out. C would equal 9.2195

4 0
3 years ago
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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