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zlopas [31]
3 years ago
8

The value of 5x2 - 11x + 18 is when x = 2.4 and whenx = -3.

Mathematics
2 answers:
Virty [35]3 years ago
8 0

Answer:

20.4

96

Step-by-step explanation:

The value of 5x² - 11x + 18 is

5(2.4)²-11(2.4) +18 = 5(5.76) -26.4+18

= 28.8 -26.4+18 = 20.4 (when x = 2.4)

and 5(-3)² -11(-3) +18 = 45+33+18 = 96

(when x = -3)

ivann1987 [24]3 years ago
4 0

f(x)=5x^2-11x+18

\\ \sf\longmapsto f(2.4)

\\ \sf\longmapsto 5(2.4)^2-11(2.4)+18

\\ \sf\longmapsto 5(5.76)-26.4+18

\\ \sf\longmapsto 28.8-26.4+18

\\ \sf\longmapsto 2.4+18

\\ \sf\longmapsto 20.4

And

\\ \sf\longmapsto f(-3)

\\ \sf\longmapsto 5(-3)^2-11(-3)+18

\\ \sf\longmapsto 5(9)+33+18

\\ \sf\longmapsto 45+51

\\ \sf\longmapsto 96

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A confidence interval (CI) is desired for the true average stray-load loss u (watts) for a certain type of induction motor when
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Answer:

A) CI = (57.12 , 59.48)

B) CI = (57.71 , 58.89)

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D) n = 239.63

Step-by-step explanation:

a)

given data:

mean, \bar X = 58.3

standard deviation, σ = 3

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ME = Zc * σ \sqrt{n}

ME = 1.96 * 3 \sqrt{25}

ME = 1.18

CI = (\bar X - Zc * s\sqrt{n}  , \barX + Zc * s\sqrt{n})

CI = (58.3 - 1.96 * 3\sqrt{25} , 58.3 + 1.96 * 3\sqrt{25})

CI = (57.12 , 59.48)

b)

Given data:

mean, \bar X = 58.3

standard deviation, σ = 3

sample size, n = 100

Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ \sqrt{n}

ME = 1.96 * 3\sqrt{100}

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CI = (58.3 - 1.96 * 3\sqrt{100} , 58.3 + 1.96 * 3\sqrt{100})

CI = (57.71 , 58.89)

c)

sample mean, \bar X = 58.3

sample standard deviation, σ = 3

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ME = Zc * σ \sqrt{n}

ME = 2.58 * 3\sqrt{100}

ME = 0.77

CI = (\bar X - Zc * s\sqrt{n}  , \barX + Zc * s\sqrt{n})

CI = (58.3 - 2.58 * 3\sqrt{100} , 58.3 + 2.58 * 3/\sqrt{100}

CI = (57.53 , 59.07)

D)

Given data:

Significance Level, α = 0.01,

Margin or Error, E = 0.5,

σ = 3

The critical value for α = 0.01 is 2.58.

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7 0
3 years ago
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