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2 years ago

In a bell-shaped curve, the x-axis (horizontal direction) of the graph represents which of the following

1 answer:

4 0

X-axis represents the value of a particular characteristic; characteristics of an organism can include such traits as size and color.

<h3>What is a bell-shaped curve?</h3>

A typical sort of distribution for a variable is the bell curve, commonly referred to as the normal distribution. The normal distribution graph, which has a symmetrical bell-shaped curve, is what gave rise to the phrase "bell curve."

The highest point on the curve, or the top of the bell, denotes the most likely outcome in a set of data (its mean, mode, and median), while all other potential outcomes are symmetrically distributed around the mean, resulting in a downward-sloping curve on each side of the peak. The bell curve's standard deviation provides information about its width.

The x-axis represents a variable's value, while the y-axis provides information about the likelihood that we will see that value.

I understand the question you are looking for is this:

In a bell-shaped curve, the x-axis (horizontal direction) of the graph represents which of the following?

a. The value of a particular characteristic; characteristics of an organism can include such traits as size and color.

b. The number of individuals

c. Time

Learn more about bell-shaped curve here:

brainly.com/question/24182600

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A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

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A 58 kg skier is going down a 35 degree slope. The areaof each

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To solve this problem we will use a free body diagram that allows us to determine the Normal Force.

In general, the normal force would be equivalent to

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