All categories

2 years ago

What percent of 134 is 66? (To the nearest tenth of a percen t)

Mathematics

1 answer:

dezoksy [38]2 years ago

3 0

Answer: 49.25

Step-by-step explanation:

You might be interested in

Susan works at the local Department store. She makes a 2% commission on all sales. She sells jewelry worth $20,000. What is her

Sergio039 [100]

Answer:

I think it's 2000

Step-by-step explanation:

4 0

3 years ago

An auto maker estimates that the mean gas mileage of its sports utility vehicle is at least 20 miles per gallon. A random sample

Nady [450]

Answer:

Yes, we reject the auto maker's claim.

Step-by-step explanation:

H0 : μ ≥ 20

H1 : μ < 20

Sample mean, xbar = 18 ;

Sample size, n = 36

Standard deviation, s = 5

At α = 0.01

The test statistic :

(xbar - μ) ÷ s /sqrt(n)

(18 - 20) ÷ 5/sqrt(36)

-2 /0.8333333

= - 2.4

Pvalue from test statistic : Pvalue = 0.00819

Pvalue < α

0.00819 < 0.01

Hence, we reject the Null

7 0

2 years ago

What is fixed perimiter

Kryger [21]

<span>This means the length of the side of an area, a rectangle, for example, is always the same.

Hope this helps!</span>

8 0

3 years ago

Read 2 more answers

4x-5y=13

love history [14]

Hello :
<span>4x-5y=13...(1)
x+5y=-3...(2)
(1)+(2) : 5x = 10
x=2
subsct in (2) : 2+5y = -3
5y =-5
y = -1

</span>

8 0

3 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

2 years ago