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Evgen [1.6K]
2 years ago
14

Please answer the following question down below. (25 points)

Mathematics
1 answer:
allsm [11]2 years ago
6 0

Answer:

It's the second answer 0.44444444%.

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The root mean square velocity of one mole of a monoatomic gas having molar mass M is U rms . The relation between average kineti
garik1379 [7]

root means square velocity of one mole of a monoatomic gas having molar mas*s m(M) is Urms. The relation between average kinetic energy (e)of the gas and Urms is Urms=M2E

StarrySoul please check it...

8 0
2 years ago
4. From 5 faculty members, a committee of 2 is to be formed. In how many ways can this be done?
ohaa [14]

Answer:

3

Step-by-step explanation:

7 0
3 years ago
A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f
Andrews [41]

Answer:

A = \frac{P}{r}\left( e^{rt} -1 \right)

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

                                                \frac{dA}{rA+P} = dt

Integration on both sides gives

                                            \int \frac{dA}{rA+P} = \int  dt

where c is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

Therefore,

                                        \frac{1}{r} \ln |rA+P| = t+c

Multiply both sides by r.

                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

Substituting 0 for A and t in the equation gives

                         0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P

Therefore, the solution of the given differential equation is

                                   A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)

4 0
3 years ago
at best buy they have a 60"tv that sells for $1999 and is on sale for 20% off. the sales tax is 8.8%. what is the final cost
sweet [91]
I believe the answer is $1,458.47
6 0
3 years ago
The measured weights of 1,000 men in a certain village follow a normal distribution with a mean of 150 pounds and a standard dev
Hitman42 [59]
Z = (X-Mean)/SD 
<span>z1 = (165 - 150)/15 = +1 </span>
<span>z2 = (135 - 150)/15 = - 1 </span>
<span>According to the Empirical Rule 68-95-99.7 </span>
<span>Mean +/- 1SD covers 68% of the values </span>
<span>100% - 68% = 32% </span>
<span>The remaining 32% is equally distributed below z = - 1 and z = +1 </span>
<span>32%/2 = 16% 
</span>
<span>Therefore, 
</span>
<span>a) Number of men weighing more than 165 pounds = 16% of 1000 = 160 </span>
<span>b) Number of men weighing less than 135 pounds = 16% of 1000 = 160</span>
3 0
3 years ago
Read 2 more answers
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