Suppose a normal distribution has a mean of 222 and a standard deviation of
16. What is the probability that a data value is between 206 and 230? Round
your answer to the nearest tenth of a percent.
A. 91.0%
B. 66.9%
C. 53.3%
D. 84.0%

Answers

Answer 1

Answer:

C) 53.3%

The probability that a data value is between 206 and 230

P( 206 ≤X≤230) = 0.5328 = 53.3%

Step-by-step explanation:

Explanation

Given that Mean of the Normal distribution(μ)  = 222

Given that the standard deviation of the Normal distribution (σ) = 16

Let 'X' be the random variable in the Normal distribution

we have to find that the probability that a data value is between 206 and 230

solution:-

Step(i):-

Let  'X' = 206

[tex]Z = \frac{x^{-}-mean }{S.D} = \frac{206-222}{16} = -1[/tex]

Let X = 230

[tex]Z = \frac{x-mean }{S.D} = \frac{230-222}{16} = 0.5[/tex]

Step(ii):-

The probability that a data value is between 206 and 230

P( 206 ≤X≤230) = P( -1≤Z≤0.5)

                          = |A(0.5)+A(-1)|

                         = 0.1915+0.3413

                         = 0.5328

final answer:-

The probability that a data value is between 206 and 230

P( 206 ≤X≤230) = 0.5328 = 53.3%


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Jayvion took out a simple interest loan to pay for some new furniture. If the loan was for 5 years at 12% and he paid $960 interest, how much money did he borrow?

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Answer:

$544.73

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Algebra I

Simple Interest Rate Formula: [tex]\displaystyle A = P(1 + r)^t[/tex]

A is final amountP is principle amountr is ratet is time

Step-by-step explanation:

Step 1: Define

A = 960

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t = 5

Step 2: Solve for P

Substitute in variables [Simple Interest Rate Formula]:                                [tex]\displaystyle 960 = P(1 + 0.12)^5[/tex][Interest] (Parenthesis) Add:                                                                            [tex]\displaystyle 960 = P(1.12)^5[/tex][Interest] Evaluate exponents:                                                                        [tex]\displaystyle 960 = P(1.76234)[/tex][Interest] [Division Property of Equality] Isolate P:                                        [tex]\displaystyle 544.73 = P[/tex][Interest] Rewrite:                                                                                            [tex]\displaystyle P = 544.73[/tex]

For the following integral, give a power or simple exponential function that if integrated on a similar infinite domain will have the same convergence or divergence behavior as the given integral, and use that to predict whether the integral converges or diverges. Note that for this problem we are not formally applying the comparison test; we are simply looking at the behavior of the integrals to build intuition.

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Answers

Answer:

hello your question is incomplete attached below is the complete question

answer :

for  I1 =     [tex]\frac{1}{x^2} + \frac{4}{x^3}[/tex]    The integral converges

for  I2 =    [tex]\frac{2x + 6}{2(x^2 + 6x + 4 )}[/tex]  The integral diverges

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Step-by-step explanation:

The similar integrands and the prediction ( conclusion )

for  I1 =     [tex]\frac{1}{x^2} + \frac{4}{x^3}[/tex]    The integral converges

for  I2 =    [tex]\frac{2x + 6}{2(x^2 + 6x + 4 )}[/tex]  The integral diverges

for  I3 =  [tex]\frac{1}{x^2}[/tex]  The integral converges

for  I4 = 1  The integral diverges

attached below is a detailed solution


The area of the four walls of a room is 156 m2. The breadth and height of the room are 8 m and 6 m
respectively. Find the length of the room.​

Answers

Given: Lateral Surface area=156m
Breadth=8m height=6m
⇒ L.S.A=2(bh+lh)=2(l+b)h
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⇒156=2(l+8)×6
⇒l+8=
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=13
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Answer:

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What distance (in miles) has the race-car traveled if the car has swept out an angle of 212 degrees?
 _____miles   

What is the measure of the angle swept out by the car (in radians) if the car has traveled 4.9 miles?
 _____radians   

Define a function, h, that gives the race-car's distance above the horizontal diameter of the track (in miles) in terms of the number of hours since the race-car started driving, t.

Answers

Answer:

The answer is below

Step-by-step explanation:

The car is moving in a circular track with a radius of 1.9 miles. The distance covered by the car if the track is revolved once = 2π * radius of the track.

a) Since the car has swept out an angle of 212 degrees, the distance covered by the race car = [tex]\frac{212}{360}*2\pi(1.9)=7.03 \ miles[/tex]

b) If the car traveled 4.9 miles, the angle swept out (θ) is:

[tex]4.9= \frac{\theta}{360} * 2\pi r\\\\\theta=\frac{4.9}{2\pi (1.9)}*360 \\\\\theta=147.76^o=147.76^o*\frac{\pi}{180} \\\\\theta=2.58\ rad[/tex]

c) h = distance covered, t = time in hourss.

Hence:

h = 85.5t

The race-car's motion round the track is a repeating motion that can be

described by a sinusoidal function.

The correct responses are;

Distance travelled when the car swept an angle of 212° ≈ 7.03 miles

If the car has travelled 4.9 miles, the angle swept out, θ ≈ 2.58 radians

The function is;  [tex]\underline{h(t) = 1.9\cdot sin(45\cdot t)}[/tex].

Reasons:

Known parameters are;

Radius of the track, r = 1.9 miles

Point the car starts = 3 O'clock

Speed of the car = 85.5 mph

Required:

Distance swept out when the car traveled an angle of 212°.

Solution;

Distance of one complete turn = 2·π·r

Angle of one complete turn = 360°

Therefore, at 212°, we have;

[tex]\dfrac{212^{\circ}}{360^{\circ}} \times 2 \times 1.9 \times \pi \approx 7.03[/tex]

Distance travelled when the car swept an angle of 212° ≈ 7.03 miles

Required:

The measure of the angle swept out by the car if the car has travelled 4.9 miles.

Solution;

Let, θ represent the angle, we have;

[tex]\dfrac{\theta}{2 \cdot \pi} \times 2 \times 1.9 \times \pi \approx 4.9 \ miles[/tex]

We get;

[tex]{\theta} = \dfrac{4.9 \ miles}{2 \times 1.9 \ miles \times \pi } \times 2\cdot \pi \approx 2.58 \ radians[/tex]

If the car has travelled 4.9 miles, the angle swept out, θ ≈ 2.58 radians

Required:

The function, h, that gives the distance of the above the horizontal

diameter of the track (in miles) in terms of the number of hours since the

race-car started driving, t.

Solution;

The function that gives the car height can be presented as follows;

The angular velocity, ω = [tex]\dfrac{v}{r}[/tex]

Therefore;

[tex]\omega = \dfrac{85.5 \ mph}{1.9 \ miles} = 45 \ rad/hour[/tex]

The general form of the sinusoidal function is h = A·sin[k·(θ - b)] + c

Therefore, we get;

h = 1.9·sin[k·(θ - b)] + c

The period, T = 2·π/ω = 2·π/k

Therefore, given that ω·t = θ, we get;

h = 1.9·sin[ω·t - b] + c = 1.9·sin[45·t - b] + c

The vertical sift, c, and the horizontal shift, b, can both be taken as zero,

given that the sin(0) = 0, therefore;

h = 1.9·sin[0] + c = c = 0, such that the at the start, t = 0, the car is at a

distance of h = 0 above the horizontal line.

The function, h, that gives the distance of the car above the

horizontal track, in terms of the number of hours since the car started

driving, t, is therefore;

[tex]\underline{h(t) = 1.9\cdot sin(45\cdot t)}[/tex].

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Answers

Give me more information please..

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Answer:

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

Write a polynomial in standard form with the zeros –4, 0, 1, and 4.

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Answer:

uhhm

Step-by-step explanation:

....

The required polynomial in standard form with zeros –4, 0, 1, and 4 is x⁴- x³ - 16x²+16x

Polynomial functions are functions that have a leading variable of 3 and above.

Given the zeros of a polynomial as -4, 0, 1, and 4, their corresponding factors will be:

(x+4)), x-0, x-1 and x - 4

To get the polynomial function required, we will take the product of the factors as shown:

= x(x-1)(x+4)(x-4)

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f(x)= x⁴-16x²-x³+16x

Rearranging

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en la librería habían 680 libros para el comienzo de clase pidieron a la editorial 125 más Cuántos libros hay ahora en total para vender​

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Answer:

Que? No se.

Step-by-step explanation:

Answer:

805

Step-by-step explanation:

As part of a project about response bias, Ellery surveyed a random sample of 25 students from her school. One of the questions in the survey required students to state their GPA aloud. Based on the responses, Ellery said she was 90% confident that the interval from 3.14 to 3.52 captures the mean GPA for all students at her school.


Required:

a. Interpret the confidence level.

b. Explain what would happen to the length of the interval if the confidence level were increased to 99%.

c. How would a 90% confidence interval based on a sample of size 200 compare to the original 90% interval?

d. Describe one potential source of bias in Ellery's study that is not accounted for by the margin of error.

Answers

Answer:

a. We are 90% sure that the mean GPA for all students at her school is between 3.14 and 3.52.

b. The length would increase.

c. The interval would be narrower.

d. The biggest potential source of bias is making the students state their grades aloud, which makes it possible that students with low grades could lie their grades, making it seem higher, and the interval biased.

Step-by-step explanation:

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

Margin of error of a confidence interval:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which z is related to the confidence level(higher the confidence level, higher z) [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

This shows that if we increase the confidence level, the margin of error gets larger and the interval gets wider, while if we increase the sample size the margin of error gets smaller and the interval gets narrower.

a. Interpret the confidence level.

We are 90% sure that the mean GPA for all students at her school is between 3.14 and 3.52.

b. Explain what would happen to the length of the interval if the confidence level were increased to 99%.

We would increase the confidence level, and the interval would get wider, that is, the length would increase.

c. How would a 90% confidence interval based on a sample of size 200 compare to the original 90% interval?

Larger sample size means that the interval would be narrower.

d. Describe one potential source of bias in Ellery's study that is not accounted for by the margin of error.

The biggest potential source of bias is making the students state their grades aloud, which makes it possible that students with low grades could lie their grades, making it seem higher, and the interval biased.

Based on the information given in the sample, it should be noted that this implies that we are 90% sure that the mean GPA for all students at the school is between 3.14 and 3.52.

SamplingThe thing that will happen to the length of the interval if the confidence level were increased to 99% is that the length would increase.

A 90% confidence interval based on a sample of size 200 compare to the original 90% interval is that the interval would be narrower.

Lastly, a potential source of bias in Ellery's study that is not accounted for by the margin of error is making the students state their grades aloud. This will make some of them with lower grades lie.

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