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3 years ago

Is 8x-2y=2 a direct or inverse variation

Mathematics

1 answer:

sp2606 [1]3 years ago

7 0

Answer:

Neither, why... for it to be direct or inverse variation, the model has to fit either y=k/x or y=kx, it may not have a y-intercept at all if it is inverse variation and it must have a y-intercept of 0 for it to be direct variation.

Step-by-step explanation:

The statement y=2 is quite specific. Because k is positive, y increases as x increases. So as x increases by 1, y increases by 1.5. Inverse Variation: Because k is positive, y decreases as x increases. yx is a constant number -8. The constant of variation, k , is 23 . Inverse Variation

An inverse variation can be represented by the equation xy=k or y=kx .

That is, y varies inversely as x if there is some nonzero constant k such that, xy=k or y=kx where x≠0,y≠0 .

Suppose y varies inversely as x such that xy=3 or y=3x . That graph of this equation shown.

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A company rounds its losses to the nearest dollar. The error on each loss is independently and uniformly distributed on [–0.5, 0

lesya [120]

Answer:

the 95th percentile for the sum of the rounding errors is 21.236

Step-by-step explanation:

Let consider X to be the rounding errors

Then; $X \sim U (a,b)$

where;

a = -0.5 and b = 0.5

Also;

Since The error on each loss is independently and uniformly distributed

Then;

$\sum X _1 \sim N ( n \mu , n \sigma^2)$

where;

n = 2000

Mean $\mu = \dfrac{a+b}{2}$

$\mu = \dfrac{-0.5+0.5}{2}$

$\mu =0$

$\sigma^2 = \dfrac{(b-a)^2}{12}$

$\sigma^2 = \dfrac{(0.5-(-0.5))^2}{12}$

$\sigma^2 = \dfrac{(0.5+0.5)^2}{12}$

$\sigma^2 = \dfrac{(1.0)^2}{12}$

$\sigma^2 = \dfrac{1}{12}$

Recall:

$\sum X _1 \sim N ( n \mu , n \sigma^2)$

$n\mu = 2000 \times 0 = 0$

$n \sigma^2 = 2000 \times \dfrac{1}{12} = \dfrac{2000}{12}$

For 95th percentile or below

$P(\overline X < 95}) = P(\dfrac{\overline X - \mu }{\sqrt{{n \sigma^2}}}< \dfrac{P_{95}- 0 } {\sqrt{\dfrac{2000}{12}}}) =0.95$

$P(Z< \dfrac{P_{95} } {\sqrt{\dfrac{2000}{12}}}) = 0.95$

$P(Z< \dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}}) = 0.95$

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1- 0.95$

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} = 0.05$

From Normal table; Z > 1.645 = 0.05

$\dfrac{P_{95}\sqrt{12} } {\sqrt{{2000}}} =1.645$

${P_{95}\sqrt{12} } = 1.645 \times {\sqrt{{2000}}}$

${P_{95} = \dfrac{1.645 \times {\sqrt{{2000}}} }{\sqrt{12} } }$

$\mathbf{P_{95} = 21.236}$

the 95th percentile for the sum of the rounding errors is 21.236

8 0

3 years ago

The square pyramid has a volume of 486 cubic inches. What is the value of x?

Softa [21]

Volume of square pyramid can be calculated by the formula :

V= $a^{2}\frac{h}{3}$

a is the length of base and h is the height of the pyramid.

In the given figure a= x in.and h = $\frac{1}{4} x$ in.

Volume of pyramid given V =486 cubic inches.

Substituting these values in volume of pyramid we have:

486= $x^{2} \frac{x}{4}$ ÷ 3

Or 486= $\frac{x^{3}}{4} x\frac{1}{3}$

486= $\frac{x^{3}}{12}$

Multiplying both sides by 12

5832 = $x^{3}$

$18^{3} =x^{3}$

Or x= 18 in.

8 0

3 years ago

Read 2 more answers

Can someone help me please?

taurus [48]

Answer:

the only answer that makes sense would be the length

4 0

3 years ago

Hey, wanna pair our account on Brainly so we can share the perks? https://brainly.com/invite/7894ac0d941dd7582da9301dad223928?ut

Anastaziya [24]

Answer:

points for me too pls

Step-by-step explanation:

4 0

3 years ago

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Can somebody please help im stuck on it :(

masha68 [24]

Answer:

Step-by-step explanation:

It is the same equation but backward

8 0

3 years ago