According to the text, the specific things that people want in life are called:

What is the probability of randomly selecting a z-score less than z = 2.25 from a normal distribution?

Vanyuwa [196]

The probability of randomly selecting a z-score less than z = 2.25 from a normal distribution is 0.8944

Probability is the branch of arithmetic concerning numerical descriptions of ways in all likelihood an occasion is to occur, or how probably it's far that a proposition is genuine. The possibility of an event is a number between zero and 1, wherein, more or less talking, zero suggests the impossibility of the event and 1 shows the truth.

Probability = the range of methods of achieving achievement. the total range of viable results. As an example, the chance of flipping a coin and it being heads is ½, due to the fact there's 1 way of having a head and the full number of possible effects is two (a head or tail). We write P(heads) = ½ .

A probability is a range that reflects the danger or chance that a particular event will occur. Chances can be expressed as proportions that vary from 0 to at least one, and they also can be expressed as probabilities ranging from zero% to 100%.

Learn more about probability here brainly.com/question/24756209

#SPJ4

4 0

2 years ago

A tuning fork with a frequency of 335 Hz and a tuning fork of unknown frequency produce beats with a frequency of 5.3 when struc

Ganezh [65]

Answer:

Explanation:

Unknown fork frequency is either

335 + 5.3 = 340.3 Hz

or

335 - 5.3 = 329.7 Hz

After we modify the known fork, the unknown fork frequency equation becomes either

(335 - x) + 8 = 340.3

(335 - x) = 332.3

x = 2.7 Hz

or

(335 - x) + 8 = 329.7

(335 - x) = 321.7

x = 13.3 Hz

IF the unknown fork frequency was 340.3 Hz,

THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz

IF the unknown fork frequency was 329.7 Hz,

THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz

3 0

3 years ago

Answer question quick

Anettt [7]

Answer:

Yes , insulation has no role

5 0

3 years ago

An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e

timama [110]

Answer:

K = 80.75 MeV

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

$E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}$

<em>where $E_{ph}$ : is the photon energy, $E_{0p} and E_{0ap}$ : are the rest energies of the proton and the antiproton, respectively, equals to m₀c², $K_{p} and K_{ap}$ : are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>

Therefore the kinetic energy of the antiproton is:

$K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}$

<u>The proton mass is equal to the antiproton mass, so</u>:

$K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV$