Answer:

True.

Explanation:

The Stanford-Binet test was developed by Alfred Binet and Theodore Simon in 1905 to measure intelligence. It was first used during World War I to identify individuals for officer status in the United States Army. The test was later revised and is still used today as an intelligence test for children and adults.

Astronomers define the __________ as all of space and everything in it. It is enormous, almost beyond imagination. Question 2 options: galaxy none of these universe solar system

**Answer:**

Universe

**Explanation:**

I took the quiz.

What is the power generated by a motor boat that applies a 1500 N force over 1000 m in 60 seconds?

The correct answer is 25 KW

The circuit has a 3 volt EMF and two ohm resistors. How much power in watts does this circuit draw? A) 4.5 , B) 24, C) 1.13 D) 2.67 E) 0.375 F) 1.5

**Answer:**

P = 4.5 watts

**Explanation:**

Given that,

EMF of the circuit, E = 3 volt

The resistance of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

[tex]P=\dfrac{V^2}{R}[/tex]

Substitute all the values,

[tex]P=\dfrac{3^2}{2}\\\\P=4.5\ W[/tex]

**So, the power of this circuit is equal to 4.5 watts.**

Which of the following is true of the deep

water layer of the ocean?

A. warmest and least dense of the ocean layers

B. experiences a rapid decrease in temperature

C. is warm in the summer and cold in the winter

D. cold all year round

The answer is D. :>

1. Although mercury is a metal, it is a liquid at room temperature. Mercury melts at about -39°C. If

the temperature of a block of mercury starts at -54°C and increases by 22°C, does the mercury melt?

Explain your answer.

**Answer:**

I think so because if it starts at a low temperature for that material, it should melt when you bring it up to that temperature.

**Answer: Mercury **

**Explanation: Mercury is kinda like the opposite of regular semi-liquid. For ice to melt you need Fahrenheit to melt ice into water while you need Celsius to melt mercury.**

A rifle fires a pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by from its unstrained length. The pellet rises to a maximum height of 6.10 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

This question is incomplete, the complete question is;

A rifle fires a 2.10 × 10⁻² kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 9.10 × 10⁻² m from its unstrained length. The pellet rises to a maximum height of 6.10 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

**Answer:**

the spring constant is **303.5 N/m**

**Explanation:**

Given the data in the question;

There is a potential energy associated with the spring;

we know that potential energy = kinetic energy

mgh = [tex]\frac{1}{2}[/tex]kx²

where k is the spring constant and x is the compression

as the pallet is fired, the spring gives kinetic energy which is converted into gravitational potential energy

so

m = 2.10 × 10⁻²

g = 9.81 m/s²

h = 6.10 m

x = 9.10 × 10⁻² m

we substitute

mgh = [tex]\frac{1}{2}[/tex]kx²

2.10 × 10⁻² × 9.81 × 6.10 = [tex]\frac{1}{2}[/tex] × K × ( 9.10 × 10⁻² )²

1.256661 = 0.0041405 × K

K = 1.256661 / 0.0041405

K = **303.5 N/m**

Therefore, the spring constant is **303.5 N/m**

A golf ball strikes a hard, smooth floor at an angle of 27.0 ° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0200 kg, and its speed is 33.0 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)

**Answer:**

J = 3.564 N.s

**Explanation:**

From the given information:

angle θ = 27°

mass = 0.0200 kg

speed = 33.0 m/s

To determine the impulse applied using the equation:

J = m(2V cos θ)

J = 0.0200 (2 × cos (27.0))

J = 0.0200 (2 × 0.8910)

J = 0.03564

J = 3.564 N.s

A researcher would like to perform an experiment in a zero magnetic field, which means that the field of the earth must be canceled. Suppose the experiment is done inside a solenoid of diameter 1.0 m, length 4.6 m , with a total of 5000 turns of wire. The solenoid is oriented to produce a field that opposes and exactly cancels the 52 Î¼T local value of the earth's field.

What current is needed in the solenoid's wires?

Express your answer to two significant figures and include the appropriate units.

**Answer:**

I = 3.81 x 10⁴ A

**Explanation:**

The magnetic field of a solenoid must be equal to the field of earth:

[tex]Field\ of\ Earth = Field\ of\ Solenoid\\52\ T = \mu n I\\I = \frac{52}{\mu n}[/tex]

where,

I = current passing through solenoid = ?

μ = permeability of free space = 4π x 10⁻⁷ N/A²

n = no. of turns per unit length = [tex]\frac{5000\ turns}{4.6\ m}[/tex] = 1086.96 /m

Therefore,

[tex]I = \frac{52\ T}{(4\pi\ x\ 10^{-7}\ N/A^2)(1086.96\ /m)}[/tex]

**I = 3.81 x 10⁴ A**

A copper wire has a mass of 29.33 mg/cm and has a length of 2.5 cm.

Find the weight of the copper wire.

The correct answer 73.325 mg

the pencil has a shadow. is the pencil opaque or transparent?

**Explanation: **

**Pencil is clearly an opaque object because, we cannot see through it. For example, take a pencil in your front. You will observe that, you are not able to see the objects behind it. And that is why you can say that pencil is an opaque object.**

**Answer:**

A pencil is opaque

**Explanation:**

A pencil is opaque, because light cannot pass through it and shadow is created.

An amusement park ride swings back and forth once every 60.0 s. What is the ride's frequency?

The Answer choices are in the image above

Answer:

Option B. 0.016 Hz

Explanation:

From the question given above, the following data were obtained.

Period (T) = 60 s

Frequency (f) =?

Frequency and period are related according to the following equation:

Frequency = 1 / period

f = 1/T

With the above formula, we can obtain the frequency of the ride as follow:

Period (T) = 60 s

Frequency (f) =?

f = 1/T

f = 1/60

f = 0.016 Hz

Thus, the rider's frequency is 0.016 Hz

What is a gravitational force?

**Answer:**

It is the force that pulls down an object on the air

**Answer: a downward pull on any object **

**Explanation: **

Consider a pulley of mass mp and radius R that has a moment of inertia 1/2mpR2. The pulley is free to rotate about a frictionless pivot at its center. A massless string is wound around the pulley and the other end of the rope is attached to a block of mass m that is initially held at rest on frictionless inclined plane that is inclined at an angle β with respect to the horizontal. The downward acceleration of gravity is g. The block is released from rest .

How long does it take the block to move a distance d down the inclined plane?

Write your answer using some or all of the following: R, m, g, d, mp,

**Answer:**

a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex] , t = [tex]\sqrt{ \frac{2d}{a} }[/tex]

**Explanation:**

To solve this exercise we must use Newton's second law

For the block

let's set a reference system with the x axis parallel to the plane

X axis

Wₓ - T = m a

Y axis

N- W_y = 0

N = W_y

for pulley

∑τ = I α

T R = (½ m_p R²) α

let's use trigonometry for the weight components

sin β = Wₓ / W

cos β = W_y / W

Wx = W sin β

angular and linear variables are related

a = α R

α = a / R

we substitute and group our equations

W sin β - T = m a

T R = ½ m_p R² (a / R)

W sin β - T = m a

T = ½ m_p a

we solve the system of equations

W sin β = (m + ½ m_p) a

a = [tex]\frac{m}{m+ \frac{1}{2} m_p} \ g \ sin \beta[/tex]

let's find the time to travel the distance (d) through the block

x = v₀ t + ½ a t²

d = 0 + ½ a t²

t = [tex]\sqrt{ \frac{2d}{a} }[/tex]

A physicist at a respected research laboratory reports a startling new

discovery. Her conclusions are based on data from many trials. However,

other scientists are unable to reproduce the results of the experiment

Which statement tells why the original experiment might be faulty?

A. The conclusion are based on multiple trials

B. The results are announced to the public

C. Experimental results cannot be produced

D. The experiment does not include sophisticated equipment

The experiment does not include sophisticated equipment

Please answer the one you know!

#8 positive kinetic energy

What must happen to a sound making object for it to make sound?

A: it must be made of metal

B: it must be in a vacuum

C: it must have force applied to it

D: it must be bent

**Answer:**

option b

**Explanation:**

........................

Lightning produces a maximum air temperature on the order of 104K, whereas a nuclear explosion produces a temperature on the order of 107K. Find the order of magnitude of the wavelength radiated with greatest intensity by each of these sources. Name the part of the EM spectrum where you would expect to radiate most strongly.

**Answer:**

tex]2.898\times 10^{-7}\ \text{m}[/tex] ultraviolet region

[tex]2.898\times 10^{-10}\ \text{m}[/tex] x-ray region

**Explanation:**

T = Temperature

b = Constant of proportionality = [tex]2.898\times 10^{-3}\ \text{m K}[/tex]

[tex]\lambda[/tex] = Wavelength

[tex]T=10^4\ \text{K}[/tex]

From Wein's law we have

[tex]\lambda=\dfrac{b}{T}\\\Rightarrow \lambda=\dfrac{2.898\times 10^{-3}}{10^4}\\\Rightarrow \lambda=2.898\times 10^{-7}\ \text{m}[/tex]

**The wavelength of the radiation will be [tex]2.898\times 10^{-7}\ \text{m}[/tex] and it is in the ultraviolet region.**

[tex]T=10^7\ \text{K}[/tex]

[tex]\lambda=\dfrac{2.898\times 10^{-3}}{10^7}\\\Rightarrow \lambda=2.898\times 10^{-10}\ \text{m}[/tex]

**The wavelength of the radiation will be [tex]2.898\times 10^{-10}\ \text{m}[/tex] and it is in the x-ray region.**

Which climate zone has hot summers, cold winters, and average amounts of precipitation throughout the year?

A. Equator

B. Polar

C. Temperate

D. Tropical

**Answer:**

i believe its temperate

**Explanation:**

The **climate zone** that has hot summers, cold winters, and average amounts of precipitation throughout the year is the **temperate climate **zone. So, option C is correct.

**Climatic zones **are defined as the **areas **that have different **climatic **conditions over the **world**.

Here,

The **climate zone** that hot summers, cold winters and average amounts of precipitation throughout the year is the **temperate **climate zone.

The **temperate **climate **zones **are the areas that have mild **temperature **condition that are located between the **subtropical **and the **polar **regions. These are zones with moderate **rainfall **that is throughout the year. The **temperature **in this zone will change greatly between the summer and winter seasons. These zones are hotter during the summer and very cold during the winter.

The temperate zone has an average **precipitation **of 800 mm.

Hence,

The **climate zone** that has hot summers, cold winters, and average amounts of precipitation throughout the year is the **temperate climate **zone.

To learn more about** climate zone**, click:

https://brainly.com/question/1080277

#SPJ2

A proton (with charge of 1.6 x 10^-19 C and mass of 1.7*10^-27 kg) traveling at a speed of 57,600,630 m/s in the + x-direction enters a region of space where there is a magnetic field of strength 0.5 T in the - z-direction. What would be the radius of the circular motion that the proton would go into if it is "trapped" in this magnetic field region?

**Answer:**

r = 1,224 10⁻² m

**Explanation:**

For this exercise let's use Newton's second law

F = m a

the force is magnetic

** F** = q **v** x **B**

The bold letters indicate vectors, the module of this expesion is

F = q v B

The direction of the force is found by the right hand rule

thumb points in the direction of the velicad + x

fingers extended in the direction of B -z

the palm is in the direction of the force + and

the acceleration of the proton is cenripetal

a = v² / r

we substitute

q v B = m v² / r

r = [tex]\frac{m \ v}{q \ B}[/tex]

let's calculate

r = [tex]\frac{1.7 \ 10^{-27} \ 5.760063 \ 10^7 }{1.6 \ 10^{-19} \ 0.5 }[/tex]

r = 1,224 10⁻² m

A pulley in the shape of a solid cylinder of mass 1.50 kg and radius 0.240 m is free to rotate around a horizontal shaft along the axis of the pulley. There is friction between the pulley and this shaft. A light, nonstretching cable is wrapped around the pulley, and the free end is tied to a 2.00 kg textbook. You release the textbook from rest a distance 0.900 m above the floor. Just before the textbook hits the floor, the angular speed of the pulley is 10.0 rad/s. What is the speed of the textbook just before it hits the floor

**Answer:**

the speed of the textbook just before it hits the floor is ** 2.4 m/s**

**Explanation:**

Given the data in the question;

mass of pulley = 1.50 kg

radius of pulley = 0.240 m

mass of text book = 2.0 kg

height from which text book was released = 0.9 m

angular speed of the pulley = 10.0 rad/s

the speed of the textbook just before it hits the floor = ?

the speed of the textbook v = angular speed of the pulley × radius of pulley

we substitute

v = 10.0 rad/s × 0.240 m

v =** 2.4 m/s**

Therefore, the speed of the textbook just before it hits the floor is ** 2.4 m/s**

a student starts his lawnmower by applying a constant tangential force of 150 N to the 0.3 kg disk-shaped flywheel. the radius of the flywheel is 18 cm. what is the flywheels angular acceleration? b. what is the angular speed of the wheel after it has turned through one revolution,( neglect friction and motor compression.) c. what is the tangent speed of a point on the rim of the flywheel?

**Answer:**

okay

**Explanation:**

please I don't know

Describe reflection and refraction. How do these processes enable astronomers to build telescopes? Do these principles work equally well for ALL types of non-optical telescopes; i.e. those built to view the non-visible regions of the electromagnetic spectrum (radio, infrared, ultraviolet, x-ray, gamma)? Explain carefully by using at least two specific examples from the list of the five.

**Answer:**

θ₁ = θ₂, n₁ sin θ₁ = n₂ sin θ₂

instruments can be built that concentrate light on a small surface

**Explanation:**

The two laws of optical geometry have been known for quite some time.

The law of reflection states that when a ray of light reaches a surface, the reflected ray comes out at the same angle as the incident ray.

θ₁ = θ₂

The law of refraction establishes how the direction of a light ray changes when it stops from one transparent medium to another.

n₁ sin θ₁ = n₂ sin θ₂

With any of these laws, instruments can be built that concentrate light on a small surface, which allows the weak light from the flares to be concentrated and objects to be measured and seen.

Electromagnetic radiation in its entire spectrum has the same properties, which is why telescopes of visible, infrared, and microwave light. Radio telescope, gamma rays use the same principles of the law of reflection and refraction. The main change between each instrument is the materials and which they are built,

a reflecting telescope must be built with a reflective surface, for example for microwaves metal surfaces are used

For gamma ray telescope it is preferred to constrict refracting telescope, therefore the material must be transparent to gamma rays

The radio telescope uses reflection and the surface is metallic, sometimes to reduce the weight the surface has holes smaller than the length of the donut that you want to measure.

PLEASE HELP! I'LL GIVE BRAINLEST

It would be option A.

Which are properties of a liquid? Check all that apply.

**Answer:**

the property of liquid are

1 they can flow from one place to another if surface is slanted

2 it cannot be compressed

Two resistors are connected in parallel. If R1 and R2 represent the resistance in Ohms (Ω) of each resistor, then the total resistance R is given by 1R=1R1+1R2. Suppose that in fact, these two resistors are actually potentiometers (resistors with variable resistance) and R1 is increasing at a rate of 0.4Ω/min and R2 is increasing at a rate of 0.6Ω/min. At what rate is R changing when R1=117Ω and R2=112Ω?

**Answer:**

1/Re= 1/R1 + 1/R2

**Explanation:**

**Two resistors **are connected in **parallel**. If **R1** and **R2** represent the **resistance **in **Ohms (Ω) **of each** resistor, **then the **total resistance R **is given by [tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]. Thus, the **rate **of **R **changes when **R₁ = 117 Ω** and

**R₂ = 112 Ω ** is **0.25 Ω/min**

For a given **resistor **connected in **parallel;**

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]

Making **R ** from the left-hand side the subject of the formula, then:

[tex]\mathbf{R = \dfrac{R_1R_2}{R_1+R_2}}[/tex]

Given that:

[tex]\mathbf{R_1 = 117,}[/tex] [tex]\mathbf{R_2 = 112 }[/tex]Now, replacing the values in the above previous equation, we have:

[tex]\mathbf{R = \dfrac{13104}{229}}[/tex]

However, the **differentiation **of **R **with respect to **time t** will give us the **rate** at which **R **is changing when** R1=117Ω **and** R2=112Ω.**

So, by **differentiating **the given equation of the **resistor **in **parallel **with respect to **time t;**

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex], we have:

[tex]\mathbf{\dfrac{1}{R^2}(\dfrac{dR}{dt})=\dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})}[/tex]

[tex]\mathbf{(\dfrac{dR}{dt})=R^2 \Bigg[ \dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=(\dfrac{13104}{229})^2 \Bigg[ \dfrac{0.4}{117^2}+\dfrac{0.6}{112^2}\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=3274.44 \Bigg[ (7.7052 \times 10^{-5} )\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=0.25\ \Omega /min}[/tex]

Therefore, we can conclude that the rate at which** R** is changing **R1=117Ω **and **R2=112Ω **is **0.25 Ω/min**

Learn more about **resistors **here:

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A standard 1kilogram weight is a cylinder 50.5mm in height and 52.0mm in diameter. What is the density of the meterial?(kg/m^3)

**Answer:**

**The correct answer is - 93.24×10^4 kg/m^3.**

**Explanation:**

Given:

height of cylinder: 50.5 mm

diameter = 52.0

then radius will be diameter/2 = 52/2 = 26

**Formula:**

Density = mass/ volume

Volume = πr^2h

solution:

Now the volume of a cylinder is v = (22/7)×r^2×h

= 22/7×26×26×50.5

= 107261.59 mm^3

Now volume in cubic meter V =10.7261 ×10^(-5) m^3

So density d = m/V = 1/(10.7261 ×10^(-5))

Or d = 93.24×10^4 kg/m^3

Vector A is 3 units length and points along the positive x-axis vector B is 4 units in length and points along negative y-axis .find magnitude and direction of the vector (a) A+B (b)A-B

**Answer:**

a) < 3 , -4 >

b) < 3 , -4 >

**Explanation:**

a) If you can imagine this, adding vectors is like putting them "tip to tail", where you put the beginning point of vector B to the end point of vector A (or vice versa). Your new vector (A+B) would be from the "tip" of vector A to the "tail" of vector B.

Mathematically, this is the same as adding the x-components of each vector together as well as the y-components.

Vector A: 3 units along the positive x-axis: < 3 , 0 >

Vector B: 4 units along the negative y-axis: < 0 , -4 >

A+B = < 3 , 0 > + < 0 , -4 > = < (3+0) , (0+(-4)) > = **< 3 , -4 >**

b) Subtracting is like adding a negative, so you could use the same "tip to tail" visual by adding the negative of vector B instead (which is B in the opposite direction).

Vector A: < 3 , 0 >

Vector B: < 0 , -4 >

Vector -B: < 0 , 4 >

A-B = A+(-B) = < 3 , 0 > + < 0 , 4 > = < (3+0) , (0+4) > = **< 3 , 4 >**

A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of 0.20 m and released from rest. Find the speed with which the object passes through its original position on the way up.

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

Which one of the following

statements about electric

current is correct?

- Current is calculated as the amount of charge that pass

a point on a circuit per time.

- Current is calculated as the amount of resistance a

charge encounters in a given time.

- Current is calculated as the amount of energy a charge

loses per unit of time.

- Current is calculated as the distance that a charge

moves per unit of time

**Answer:**

Current is calculated as the amount of charge that pass a point on a circuit per time.

**Explanation:**

Electric current is defined as the amount of charge that passes a point on a circuit per unit time. The mathematical definition of electric current is given by :

[tex]I=\dfrac{q}{t}[/tex]

Where

q is a charge (Coulomb)

t is time (in seconds)

**The SI unit of electric current is A. It is equivalent to C/s. So, the correct option is (A).**

A uniform electric field is present in the region between infinite parallel plane plates A and B and a uniform electric field is present in the region between infinite parallel plane plates B and C. When the plates are vertical, is directed to the right and to the left. The signs of the charges on plates A, B and C may be:_______.

A. -- ,--, --

B. + , -- , --

C. + , --, +

**Answer:**

the correct answer is C +, - , +

**Explanation:**

The electric field for positive charges is outgoing and for negative charges it is directed towards the charge.

Let's apply this to our case:

On plates A and B the field goes to the right, therefore plate A must be positive

In plates B and C the way it goes to the left, so the field is selected from it, this implies that plate C is positive

therefore plate B must be negative for both cases

when checking the correct answer is C +, - , +

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