Please help me fasttttttttt

Cathie made 35 copies of a 12-page storybook. How many copies did Cathie Make in all?

Nostrana [21]

Answer:

420 pages

Step-by-step explanation:

Take the number of copies in 1 book and multiply by the number of books

12*35 =420 pages

5 0

3 years ago

Evaluate the integral Integral from (5 comma 2 comma 4 )to (7 comma 9 comma negative 3 )y dx plus x dy plus 3 dz by finding para

padilas [110]

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_{(5,2,4)}^{(7,9,-3)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz$

Parameterize the line segment (call it $C$ ) by

$\vec r(t)=(1-t)(5\,\vec\imath+2\,\vec\jmath+4\,\vec k)+t(7\,\vec\imath+9\,\vec\jmath-3\,\vec k)$

$\vec r(t)=(2t+5)\,\vec\imath+(7t+2)\,\vec\jmath+(4-7t)\,\vec k$

with $0\le t\le1$ . Then

$\vec r'(t)=2\,\vec\imath+7\,\vec\jmath-7\,\vec k$

and the line integral is

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^1((7t+2)\,\vec\imath+(2t+5)\,\vec\jmath+3\,\vec k)\cdot\vec r'(t)\,\mathrm dt$

$=\displaystyle\int_0^1((2t+5)+2(7t+2)-21)\,\mathrm dt$

$=\displaystyle\int_0^1(28t+18)\,\mathrm dt=\boxed{32}$

Alternatively, if we can show that $\vec F$ is conservative, then we can apply the fundamental theorem of calculus. We need to find $f$ such that $\nabla f=\vec F$ , which requires

$\dfrac{\partial f}{\partial x}=y$

$\dfrac{\partial f}{\partial y}=x$

$\dfrac{\partial f}{\partial z}=3$

Integrating both sides of the first equation with respect to $x$ gives

$f(x,y,z)=xy+g(y,z)$

Differentiating both sides wrt $y$ gives

$\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}$

$\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

Differentiating wrt $z$ gives

$\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}$

$\implies h(z)=3z+C$

So we have

$f(x,y,z)=xy+3z+C$

and $\vec F$ is conservative. By the FTC, we find

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=f(7,9,-3)-f(5,2,4)=\boxed{32}$

5 0

3 years ago

At an angle of elevation 610

Vadim26 [7]

Answer:

The height of the cliff CD is approximately 539.76 m

Step-by-step explanation:

The given parameters are;

The first angle of elevation with which the captain sees the person on the cliff = 61°

The second angle of elevation with which the captain sees the person on the cliff after moving 92 m closer to the cliff = 69°

The angle made by the adjacent supplementary angle to the second angle of elevation = 180° - 69° = 111°

∴ Whereby, the rays from the first and second angle of elevation and the distance the ship moves closer to the cliff forms an imaginary triangle, we have;

The angle in the imaginary triangle subtended by the distance the ship moves closer to the cliff = 180° - 111° - 61° = 8°

By sine rule, we have;

AB/(sin(a)) = BC/(sin(c))

Which gives;

92/(sin(8°)) = BC/(sin(61°))

BC = (sin(61°)) × 92/(sin(8°)) ≈ 578.165 m

BC ≈ 578.165 m

The height CD = BC × sin(69°)

∴ The height of the cliff CD = 578.165 m × sin(69°) ≈ 539.76 m.

The height of the cliff CD ≈ 539.76 m.

5 0

3 years ago

How do you solve this

34kurt

You add 1.7 and 3.1 and that will be the answer to r which is 2.6

8 0

3 years ago

The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confide

kondor19780726 [428]

Answer:

99% confidence interval for the population standard deviation = (74.97 , 635.20).

Step-by-step explanation:

We are given that the ages of a random sample of five university professors are 39, 54, 61, 72 and 59. Also, it is provided that the ages of university professors are normally distributed.

So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation

$\sigma$ = population standard deviation

n = sample of university professors = 5

Also, $s^{2}$ = $\frac{\sum (X-\bar X)^{2} }{n-1}$ = 144.5

So, 99% confidence interval for population standard deviation, $\sigma$ is;

P(0.2070 < $\chi^{2} __5_-_1$ < 14.86) = 0.99 {As the table of $\chi^{2}$ at 4 degree of freedom

gives critical values of 0.2070 & 14.86}

P(0.2070 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 14.86) = 0.99

P( $\frac{ 0.2070}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{ 14.86}{(n-1)s^{2} }$ ) = 0.99

P( $\frac{ (n-1)s^{2}}{14.86 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{0.2070 }$ ) = 0.99

99% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{14.86 }$ , $\frac{ (n-1)s^{2}}{0.2070 }$ )

= ( $\frac{ (5-1) \times 144.5^{2}}{14.86 }$ , $\frac{ (5-1) \times 144.5^{2}}{0.2070 }$ )

= (5620.525 , 403483.092)

99% confidence interval for $\sigma$ = ( $\sqrt{5620.525}$ , $\sqrt{403483.092}$ )

= (74.97 , 635.20)

Therefore, 99% confidence interval for the population standard deviation of the ages of all professors at the university is (74.97 , 635.20).

8 0

4 years ago