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1 year ago

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Which function has no horizontal asymptote? F (x) = StartFraction 2 x minus 1 Over 3 x squared EndFraction f (x) = StartFraction

x minus 1 Over 3 x EndFraction f (x) = StartFraction 2 x squared Over 3 x minus 1 EndFraction f (x) = StartFraction 3 x squared Over x squared minus 1 EndFraction

2 answers:

Oduvanchick [21]1 year ago

8 0

Answer:

f(x)=\frac{2x^2}{3x-1}

Step-by-step explanation:

The function has a horizontal asymptote which is

The function has a horizontal asymptote which is

The function has no horizontal asymptote because the numerator has a degree which is higher than the degree of the denominator,

yanalaym [24]1 year ago

3 0

Answer:

$f(x)=\frac{2x^2}{3x-1}$

Step-by-step explanation:

The function $f(x)=\frac{2x-1}{3x}$ has a horizontal asymptote which is $y=\frac{2}{3}$

The function $f(x)=\frac{x-1}{3x}$ has a horizontal asymptote which is $y=\frac{1}{3}$

The function $f(x)=\frac{2x^2}{3x-1}$ has no horizontal asymptote because the numerator has a degree which is higher than the degree of the denominator,

The function $f(x)=\frac{3x^2}{x^2-1}$ has a horizontal asymptote which is $y=3$

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Four students, Jackson, Chloe, Lydia, and Noah, line up one behind the other. How

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4^4 = 256

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1 year ago

The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 88/(2 + x2 + y2), where T is measured in °C and x,

-Dominant- [34]

Answer:

$D_uT(3,1)=-\frac{44}{9}*\frac{1}{\sqrt{2} } \approx-3.46$

Step-by-step explanation:

To find the rate of change of temperature with respect to distance at the point (3, 1) in the x-direction and the y-direction we need to find the Directional Derivative of T(x,y). The definition of the directional derivative is given by:

$D_uT(x,y)=T_x(x,y)i+T_y(x,y)j$

Where i and j are the rectangular components of a unit vector. In this case, the problem don't give us additional information, so let's asume:

$i=\frac{1}{\sqrt{2} }$

$j=\frac{1}{\sqrt{2} }$

So, we need to find the partial derivative with respect to x and y:

In order to do the things easier let's make the next substitution:

$u=2+x^2+y^2$

and express T(x,y) as:

$T(x,y)=88*u^{-1}$

The partial derivative with respect to x is:

Using the chain rule:

$\frac{\partial u}{\partial x}=2x$

Hence:

$T_x(x,y)=88*(u^{-2})*\frac{\partial u}{\partial x}$

Symplying the expression and replacing the value of u:

$T_x(x,y)=\frac{-176x}{(2+x^2+y^2)^2}$

The partial derivative with respect to y is:

Using the chain rule:

$\frac{\partial u}{\partial y}=2y$

Hence:

$T_y(x,y)=88*(u^{-2})*\frac{\partial u}{\partial y}$

Symplying the expression and replacing the value of u:

$T_y(x,y)=\frac{-176y}{(2+x^2+y^2)^2}$

Therefore:

$D_uT(x,y)=(\frac{1}{\sqrt{2} } )*(\frac{-176x}{(2+x^2+y^2)^2} -\frac{176y}{(2+x^2+y^2)^2})$

Evaluating the point (3,1)

$D_uT(3,1)=(\frac{1}{\sqrt{2} } )*(\frac{-176(3)-176(1)}{(2+3^2+1^2)^2})=(\frac{1}{\sqrt{2} })* (-\frac{704}{144})=(\frac{1}{\sqrt{2} }) ( - \frac{44}{9})\approx -3.46$

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1 year ago

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2 years ago

It is claimed that 55% of marriages in the state of California end in divorce within the first 15 years. A large study was start

kykrilka [37]

Answer:

0.0045 = 0.45% probability that less than two of them ended in a divorce

Step-by-step explanation:

For each marriage, there are only two possible outcomes. Either it ended in divorce, or it did not. The probability of a marriage ending in divorce is independent of any other marriage. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

55% of marriages in the state of California end in divorce within the first 15 years.

This means that $p = 0.55$

Suppose 10 marriages are randomly selected.

This means that $n = 10$

What is the probability that less than two of them ended in a divorce?

This is

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.55)^{0}.(0.45)^{10} = 0.0003$

$P(X = 1) = C_{10,1}.(0.55)^{1}.(0.45)^{9} = 0.0042$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0003 + 0.0042 = 0.0045$

0.0045 = 0.45% probability that less than two of them ended in a divorce

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1 year ago

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mezya [45]

P(82 - q < x < 82 + q) = 0.44
P(x < 82 + q) - P(82 - q) = 0.44
P(z < (82 + q - 82)/7.4 - P(z < (82 - q - 82)/7.4) = 0.44
P(z < q/7.4) - P(z < -q/7.4) = 0.44
P(z < q/7.4) - (1 - P(z < q/7.4) = 0.44
P(z < q/7.4) - 1 + P(z < q/7.4) = 0.44
2P(z < q/7.4) - 1 = 0.44
2P(z < q/7.4) = 1.44
P(z < q/7.4) = 0.72
P(z < q/7.4) = P(z < 0.583)
q/7.4 = 0.583
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8 0

1 year ago