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inna [77]
1 year ago
11

Find the area under the standard normal curve to the right of z=0.49. round your answer to four decimal places, if necessary.

Physics
1 answer:
VikaD [51]1 year ago
4 0

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

<h3>What is  the standard normal curve?</h3>

The horizontal axis is approached by the standard normal bend as it extends indefinitely both in directions without ever being touched by it. The center of the bell-shaped, z=0 standard normal curve. Between z=3 and z=3, almost the entire area underneath the standard normal curve is located.

<h3>Use of the standard normal curve:</h3>

Use the normal distribution's standard form to calculate probability. Since the standard normal distribution is indeed a probability distribution, the probability that a variable will take on a range of values is indicated by area of the curve between two points. 100% or 1 is the total area beneath the curve.

<h3>According to the given data:</h3>

the region to the left of the standard normal curve,

z=0.49

To the right of,

z = 2.05

So,

The area will be:

= P[z < 0.49] + P[ z >2.05]

= P[z < 0.49] + 1 -  P[ z < 2.05]

= .6879 + 1 - .9798

= 0.7081

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

To know more about standard normal curve visit:

brainly.com/question/12972781

#SPJ4

I understand that the question you are looking for is:

Find the area under the standard normal curve to the left of z = 0.49 and to the right of z = 2.05. Round your answer to four decimal places, if necessary.

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Answer:

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\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)

Solving for i

\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A

i=2.28\ A

The potential difference across the headlight  bulbs is

V=\varepsilon  -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V

V=11.86\ V

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

V=\varepsilon  -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V

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If the temperature of water warming on a hot plate rises 10 degrees Celsius [°C], what is the change in temperature in units of
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Answer:

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°F = 9/5*(10)

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A ball thrown vertically upward is caught by the thrower after 2.93 s. Find the initial velocity of the ball. The acceleration o
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Answer:

The initial velocity of the ball is 28.714 m/s

Explanation:

Given;

time of flight of the ball, t = 2.93 s

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initial velocity of the ball, u = ?

The initial velocity of the ball is given by;

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