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1 year ago

Determine whether the underlined numerical value is a parameter or a statistic. explain your reasoning. a certain zoo

Physics

1 answer:

irinina [24]1 year ago

6 0

Parameter, because the data set of all 843 animals in a zoo is a population.

The population of all animals in the Zoo is 843

Proportion = 8%, which is the ratio regarding the population. So 8% is populaton proporiton, i..e Parameter .

A zoo, short for the zoological garden, is a place where animals are kept in enclosures, cared for, presented to the public, and occasionally bred for conservation efforts. The phrase zoological garden refers to zoology, the study of animals. It is also used to refer to an animal park or menagerie. In the United States alone, over 181 million people visit zoos every year. The goal of the modern zoo, which first appeared in the 19th century in the United Kingdom was to entertain and inspire the public by offering scientific inquiry and later educational exhibits. The London Zoological Gardens, which was opened to the public in 1847 after being opened for scientific study in 1828, is where the abbreviation "zoo" first appeared.

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A film badge worn by a radiologist indicates that she has received an absorbed dose of 2. 5 x 10-6 gy. the mass of the radiologi

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Energy absorbed by the radiologist is 1.775*10^-6J.

To find the answer, we have to know about the radiation.

<h3>How much energy has she absorbed?</h3>

We have the expression for dose of absorption as,

$D=\frac{E}{m}$

where; E is the energy absorbed and m is the mass of the body.

From the above expression, substituting appropriate values given in the question, we get,

$E=D*m=2.5*10^{-6}J/kg*71kg=1.775*10^{-4}J$

Thus, we can conclude that, the Energy absorbed by the radiologist is 1.775*10^-6J.

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1 year ago

A hiker, caught in a rainstorm might absorb 1 liter of water in her clothing. if it is windy so that this amount of water is eva

NeTakaya

Water evaporates at 100⁰C

So change in temperature = 100-20 = 80⁰C

Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg

Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg

So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ

So amount of heat require to evaporate water = 334.88 kJ

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2 years ago

What is the force of gravity between two 50.0kg masses that are separated by 0.300m?3.71x10-8N5.59x10-7N2.78x104N1.85x10-6N

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We will have the following:

$\begin{gathered} F=G\frac{m_1m_2}{r^2}\Rightarrow F=\frac{(6.67\ast10^{-11}m^3\ast kg^{-1}\ast s^{-2})(50kg)(50kg)}{(0.3m)^2} \\ \\ \Rightarrow F=1.852777778...\ast10^{-6}N\Rightarrow F\approx1.85\ast10^{-6}N \end{gathered}$

So, the force is approximately 1.85*10^-6 N.

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7 months ago

The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n

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Answer:

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Explanation:

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2 years ago

1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.

marta [7]

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

$B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB$

<h3>Speed of sound at the given temperature</h3>

$v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s$

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

$f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )$

where;

-v₀ is velocity of the observer moving away from the source
-vs is the velocity of the source moving towards the observer
fs is the source frequency
fo is the observed frequency
v is speed of sound

$f_0 = f_s(\frac{v-v_0}{v- v_s} )$

$f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz$

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1 year ago