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nadezda [96]
3 years ago
12

Analysis of the relationship between the fuel economy​ (mpg) and engine size​ (liters) for 35 models of cars produces the regres

sion model ModifyingAbove mpg with caret equals 36.44 minus 3.829 times Engine size. If a car has a 5 liter​ engine, what does this model suggest the gas mileage would​ be?
Physics
1 answer:
kvasek [131]3 years ago
8 0

Answer:

Gas mileage is 17.29

Explanation:

Given data:

The total number of the model is 35

The total size of the engine is 5 ltr

The regression model is given as

36.44 - 3.829\times engine\ size

From the information given in question we have

Regression equation is : model- mpg = 36.44 - 3.829\times engine\ size

Therefore for engine capacity of 5 liters;

Gas mileage = 36.44 - 3.829\times 5  = 17.29

Gas mileage is 17.29

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A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so
joja [24]

ANSWER:100dB

f<em>rom the  sound intensity level,the sound intensity is calculated as:</em>

<em />\beta<em>₁=</em>\beta<em>=(10dB)㏒₁₀(l₁/l₀)</em>

<em>inserting numbers:</em>

<em>120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²</em>

<em>Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which </em>

<em>can be used to solve for l₁ and get</em>

<em>          l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²</em>

<em>since the sound  intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in</em>

<em>ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:</em>

<em>l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²</em>

<em>since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:</em>

<em>     </em>\beta<em>₂=</em>\beta<em>=(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)</em>

<em>    </em>\beta<em>₂=(10dB)㏒₁₀(2.04×10¹⁰)</em>

<em />\beta<em> =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB</em>

<em />

3 0
4 years ago
A container contains 26 ounces of salt. Convert 26 ounces to pounds.
Reika [66]

Answer:

1.625 Pounds Explanation:

5 0
3 years ago
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Suppose the first coil of wire illustrated in the simulation is wound around the iron core 7 times. Suppose the second coil is w
SVETLANKA909090 [29]

Answer:

Emf induced in the coli will be equal to 26 volt

Explanation:

We have given number of turns N = 13

It is given that magnetic flux changes from 2.1 Weber to 2.7 Weber in 0.3 sec

Change in flux d\Phi =2.7-2.1=0.6Weber

Change in time dt = 0.3 sec

We have to find the induced emf

Induced emf is equal to e=N\frac{d\Phi }{dt}=13\times \frac{0.6}{0.3}=26volt

So emf induced in the coil will be equal to 6 volt

5 0
3 years ago
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Help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee I am going to follow youuu
Y_Kistochka [10]

Answer:

Answer 11

Explanation:

. Newton's law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them

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3 years ago
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The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




7 0
3 years ago
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