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1 year ago

12

solomon has 4 hats and wears a different one everyday throughout 4 days. how many different orders can he wear the hats during t

hose days?

1 answer:

Vanyuwa [196]1 year ago

5 0

4!= 24 ways
........

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Simplify 4a^3b^2 x 3a*6b^5

Strike441 [17]

Answer:

Step-by-step explanation:

(4a³b² )* (3a⁶b⁵) = 4*3 *a⁽³⁺⁶⁾ * b⁽²⁺⁵⁾

= 12a⁹b⁷

4 0

1 year ago

A potter works 4 days a week, makes 14 pots per day on average, and charges $24 a pot. If she lowers her price to $16 a pot, by

iragen [17]

Hello!

Your answer is 7.

Toodles~

5 0

1 year ago

Read 2 more answers

A tree has 99,400 leaves before autumn. Once autumn begins, the number of leaves on a tree decreases at a rate of 13% per day. A

SCORPION-xisa [38]

The answer to this question would be: D.99,400 * (0.87)^t < 12,000; 16 days
In this question, the initial number of leaves is 99,400 and it will decrease by 13% each day. Decrease by 13% can be put become: 100%-13%=87% of the original number.

Then, the function for the number of the leaves should be:
99,400 * (0.87)^t

to find how many days after autumn that the tree leaves is less than 12,000 the calculation should be:
99,400* (0.87)^t <12,000
(0.87)^t< 0.1207

The smallest number would be (0.87)^16= 0.10772290133

6 0

2 years ago

Read 2 more answers

On average, indoor cats live to 16 years old with a standard deviation of 2.5 years. Suppose that the distribution is normal. Le

Y_Kistochka [10]

Answer:

(a) N (16, 2.5²)

(b) 0.241

(c) Low: 15.4 years

High: 16.6 years

Step-by-step explanation:

The random variable <em>X</em> is defined as the age at death of a randomly selected indoor cat.

(a)

The distribution of X is:

$X\sim N(\mu = 16, \sigma^{2}=2.5^{2})$

(b)

Compute the probability that an indoor cat dies when it is between 17.2 and 19.6 years old as follows:

$P(17.2$

$=P(0.48$

Thus, the probability that an indoor cat dies when it is between 17.2 and 19.6 years old is 0.241.

(c)

Compute the two numbers within which 20% of indoor cats' age of death lies as follows:

$P(x_{1}$

The corresponding value of <em>z</em> is, 0.25.

Compute the value of <em>x</em>₁ and <em>x</em>₂ as follows:

$z=\frac{x_{1}-\mu}{\sigma}\\\\0.25=\frac{x_{1}-16}{2.5}\\\\x_{1}=16+(0.25\times 2.5}\\\\x_{1}=16.625\\\\x_{1}\approx 16.6$ $z=\frac{x_{1}-\mu}{\sigma}\\\\-0.25=\frac{x_{2}-16}{2.5}\\\\x_{2}=16-(0.25\times 2.5}\\\\x_{2}=15.375\\\\x_{2}\approx 15.4$

Low: <u>15.4</u> years

High: <u>16.6</u> years

8 0

1 year ago

Evaluate the surface integral s f · ds for the given vector field f and the oriented surface s. in other words, find the flux of

Flura [38]

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+10\,\mathbf k$
$\implies\nabla\cdot\mathbf f(x,y,z)=1+1+0=2$

By the divergence theorem, the surface integral along $S$ is equivalent to the triple integral over the region $R$ bounded by $S$ :

$\displaystyle\iint_S\mathbf f(x,y,z)\,\mathrm dS=\iiint_R\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=2\iiint_R\mathrm dV$

Convert to cylindrical coordinates, setting

$\begin{cases}x=r\cos\theta\\y=Y\\z=r\sin\theta\end{cases}\implies\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dY$

The triple integral is then equivalent to

$=\displaystyle2\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{Y=0}^{Y=2-r\cos\theta}r\,\mathrm dY\,\mathrm dr\,\mathrm\theta$
$=\displaystyle2\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r(2-r\cos\theta)\,\mathrm dr\,\mathrm\theta$
$=\displaystyle\frac23\int_{\theta=0}^{\theta=2\pi}(3-\cos\theta)\,\mathrm dr\,\mathrm\theta$
$=4\pi$

6 0

2 years ago