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2 years ago

Jacob earns $19.00 for walking 4 dogs. How much money would he earn for walking 7 dogs? Walking 8 dogs? 1 dog?

Mathematics

1 answer:

VMariaS [17]2 years ago

7 0

Answer:

$19.00 = 4x

Divide 4 on both sides

4.75 = x He ears $4.75 for one dog

$4.75(x)

$4.75(7 dogs)

= $33.25

--------------

$4.75(x)

$4.75(8 dogs)

= $38

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Slader records show that the average life expectancy of a pair of shoes is 2.2 years with astandard deviation of 1.7 years. A ma

Oksanka [162]

Answer:

For every 1000 pairs sold, the manufacturer expect to replace 239 pairs for free.

Step-by-step explanation:

Given:

Mean (μ) = 2.2, Standard deviation(S.D) (σ) = 1.7 years and x = 1 (1 year)

Let's find the Z score.

Z = $\frac{x - mean}{S.D}$

Now plug in the given values in the above formula, we get

Z = $\frac{1 - 2.2}{1.7} = -0.71$

Now we have to use the z-score table.

The z-score for 0.71 is 0.2611

Since it z is negative, so we subtract 0.2611 from 0.5000

0.5000 - 0.2611 = 0.2389

Percentage = 0.2389 × 100 = 23.89%

To find replaces for 1000 pairs, we need to multiply 23.89% by 1000

= $\frac{23.89}{100} .1000 = 238.9$

= 239

The cannot be in decimal, when we round off to the nearest whole, we get

239

3 0

3 years ago

Consider the equation y = –3(x – 3) Which equation, when graphed with the given equation, will form a system with lines that ove

pentagon [3]

Answer:

y=-3x+9 (D)

Step-by-step explanation:

8 0

3 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$