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3 years ago

Write and simplify the equation you would use to solve the problem below.

Mathematics

2 answers:

Pavel [41]3 years ago

5 0

Six shirts
77-5=72
72/12=6

aniked [119]3 years ago

5 0

($77 - $5) / ($12) = number of shirts
($72) / ($12) = number of shirts
6 = number of shirts

The answer is above in bold
Hope it helped :)

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assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. find the

drek231 [11]

Answer:

65.3658 inches

Step-by-step explanation:

Let X be the height of a woman randomly choosen. We know tha X have a mean of 63.6 inches and a standard deviation of 2.5 inches. For an x value, the related z-score is given by z = (x-63.6)/2.5. We are looking for a value $x_{0}$ such that $P(X < x_{0}) = 0.76$ , but, $0.76 = P(X < x_{0}) = P((X-63.6)/2.5 < (x_{0}-63.6)/2.5) = P(Z < (x_{0}-63.6)/2.5)$ , i.e., $(x_{0}-63.6)/2.5$ is the 76th percentile of the standard normal distribution. So, $(x_{0}-63.6)/2.5 = 0.7063$ , $x_{0} =63.6+(2.5)(0.7063) = 65.3658$ . Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.

7 0

3 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

3 years ago

Paraphin [41]

<h3> Hope this attachment helps u</h3>

5 0

3 years ago

PLEASE HELP

Gekata [30.6K]

Answer:

a. n x (2n +1) = 10

b. length: 5 width: 2

Step-by-step explanation:

5 is 1 more then 2 times 2

7 0

3 years ago

Evaluate 7(m-11) if m =11

melamori03 [73]

Answer:

Step-by-step explanation:

Plug in 11 for m. Remember to follow PEMDAS. First solve the parenthesis, then multiply.

m = 11

7(m - 11) = 7(11 - 11) = 7(0) = 0

0 is your answer.

7 0

3 years ago

Read 2 more answers