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2 years ago

Which equation is the inverse of y = 16x2 + 1?

Mathematics

1 answer:

solniwko [45]2 years ago

6 0

Step-by-step explanation:

$y = 16 {x}^{2} + 1 \\ \\ y - 1 = 16 {x}^{2} \\ \\ \frac{y - 1}{16} = {x}^{2} \\ \\ \pm \sqrt{ \frac{y - 1}{16} } = x \\ \\ x = \pm \frac{ \sqrt{y - 1} }{4} \\ \: this \: is \: the \: required \: inverse \: \\ equation \: of \: y = 16 {x}^{2} + 1$

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aliina [53]

Answer:

Step-by-step explanation:

Hello!

X₁: speed of a motorcycle at a certain intersection.

n₁= 135

X[bar]₁= 33.99 km/h

S₁= 4.02 km/h

X₂: speed of a car at a certain intersection.

n₂= 42 cars

X[bar]₂= 26.56 km/h

S₂= 2.45 km/h

Assuming

X₁~N(μ₁; σ₁²)

X₂~N(μ₂; σ₂²)

and σ₁² = σ₂²

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The parameter of interest is μ₁-μ₂

(X[bar]₁-X[bar]₂)± $t_{n_1+n_2-2}$ * $Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$

$t_{n_1+n_2-2;1-\alpha /2}= t_{175; 0.95}= 1.654$

$Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{134*16.1604+41*6.0025}{135+42-2} } = 3.71$

[(33.99-26.56) ± 1.654 *( $3.71*\sqrt{\frac{1}{135} +\frac{1}{42} }$ )]

[6.345; 8.514]= [6.35; 8.51]km/h

<em>Construct the 98% confidence interval for the difference μ₁-μ₂ when X[bar]₁= 475.12, S₁= 43.48, X[bar]₂= 321.34, S₂= 21.60, n₁= 12, n₂= 15</em>

$t_{n_1+n_2-2;1-\alpha /2}= t_{25; 0.99}= 2.485$

$Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{11*(43.48)^2+14*(21.60)^2}{12+15-2} } = 33.06$

[(475.12-321.34) ± 2.485 *( $33.06*\sqrt{\frac{1}{12} +\frac{1}{15} }$ )]

[121.96; 185.60]

I hope this helps!

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2 years ago

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