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Mathematics

1 answer:

Readme [11.4K]3 years ago

6 0

Answer:

1) y= 2x-4

2) y= 2/3 - 3

3) y=1/2x+2

4) y=-4x-4

Step-by-step explanation:

all you basically has to do is is subtract the x to isolate the y and if the y and if the y still has a number by it you use it to decide it

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The equation t^3=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

ankoles [38]

Answer:

2√2

Step-by-step explanation:

We can find the relationship of interest by solving the given equation for A, the mean distance.

<h3>Solve for A</h3>

$T^3=A^2\\\\A=\sqrt{T^3}=T\sqrt{T}\qquad\text{take the square root}$

<h3>Substitute values</h3>

The mean distance of planet X is found in terms of its period to be ...

$D_x=T_x\sqrt{T_x}$

The mean distance of planet Y can be found using the given relation ...

$T_y=2T_x\\\\D_y=T_y\sqrt{T_y}=2T_x\sqrt{2T_x}=(2\sqrt{2})T_x\sqrt{T_x}\\\\D_y=2\sqrt{2}\cdot D_x$

The mean distance of planet Y is increased from that of planet X by the factor ...

2√2

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2 years ago

Complete parts (a) and (b) using the probability distribution below.

katen-ka-za [31]

Μ = (0×0.026) + (1×0.072) +(2×0.152) + (3×0.303) + (4×0.215) + (5×0.164) + (6×0.066)
μ = 0 + 0.072 + 0.304 + 0.909 + 0.86 + 0.82 + 0.396
μ = 3.361 ≈ 3.4

We need the value of ∑X² to work out the variance
∑X² = (0²×0.026) + (1²×0.072) + (2²×0.152) + (3²×0.303) + (4²×0.215) + (5²×0.164) + (6²×0.066)
∑X² = 0+0.072+0.608+2.727+3.44+4.1+2.376
∑X² = 13.323

Variance = ∑X² - μ²
Variance = 13.323 - (3.4)² = 1.763 ≈ 2

Standard Deviation = √Variance = √1.8 = 1.3416... ≈ 1.4

The correct answer related to the value of mean and standard deviation is the option D
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3 years ago

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Step-by-step explanation:

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3 years ago

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