Let's just change your problem to 147 of 30%

"Of" is always equaling to multiplication (×)

Now, it would be easier to simplify

30% (147)
30% ÷ 100 = 0.3
0.3(147)
Now, we get to your answer which might be 44.1

4 0

3 years ago

During a 35% off sale at the furniture store, you found a chair with the original price of $84.61. How much will the sale price

Dafna11 [192]

Answer:55$

Step-by-step explanation:84.61 x .35 = 29.61

84.61 - 29.61 = 55

7 0

3 years ago

What is a quartile of 35, 45, 50, 65,85

Georgia [21]

Answer:

Lower quartile: 21.5

Median: 45

Upper quartile: 57.5

Step-by-step explanation:

pls mark brainliest

7 0

3 years ago

evaluate the fermi function for an energy KT above the fermi energy. find the temperature at which there is a 1% probability tha

dybincka [34]

Complete Question

Evaluate the Fermi function for an energy kT above the Fermi energy. Find the temperature at which there is a 1% probability that a state, with an energy 0.5 eV above the Fermi energy, will be occupied by an electron.

Answer:

The Fermi function for the energy KT is $F(E_o) = 0.2689$

The temperature is $T_k = 1261 \ K$

Step-by-step explanation:

From the question we are told that

The energy considered is $E = 0.5 eV$

Generally the Fermi function is mathematically represented as

$F(E_o) = \frac{1}{e^{\frac{[E_o - E_F]}{KT} } + 1 }$

Here K is the Boltzmann constant with value $k = 1.380649 *10^{-23} J/K$

$E_F$ is the Fermi energy

$E_o$ is the initial energy level which is mathematically represented as

$E_o = E_F + KT$

$F(E_o) = \frac{1}{e^{\frac{[[E_F + KT] - E_F]}{KT} } + 1}$

=> $F(E_o) = \frac{1}{e^{\frac{KT}{KT} } + 1}$

=> $F(E_o) = \frac{1}{e^{ 1 } + 1}$

=> $F(E_o) = 0.2689$

Generally the probability that a state, with an energy 0.5 eV above the Fermi energy, will be occupied by an electron is mathematically represented by the Fermi function as

$F(E_k) = \frac{1}{e^{\frac{[E_k - E_F]}{KT_k} } + 1 } = 0.01$