X² – 5x +3=0

Suppose that IQ scores have a bell-shaped distribution with a mean of 101 and a standard deviation of 12. Using the empirical ru

7nadin3 [17]

95.44% IQ scores are between 77 and 125

We're given,

Mean (a)=101

standard deviation (b)=12

To find : $P[77 < x < 125]$

Using Empirical Rule:

∴ $P[77 < x < 125]$ $=$ $P[\frac{77-101}{12} < \frac{x-a}{b} < \frac{125-101}{12}]\\$

$=P[-2 < z < 2]\\$

$=P[Z < 2]-P[Z < -2]\\$

$=0.9772-0.0228\\$

$=0.9544$ =95.44% (approx)

Learn more about Empirical rule of IQ calculation here:

brainly.com/question/13077017

#SPJ10

3 0

2 years ago

Pat needs to determine the height of a tree before cutting it down to be sure that it will not fall on a nearby fence. the angle

Mekhanik [1.2K]

The height of the tree, which Pat needs to determine, before cutting it down to be sure that it will not fall on a nearby fence is 111 ft.

<h3>What is right angle triangle property?</h3>

In a right angle triangle, the ratio of the opposite side to the base side is equal to the tangent angle between them.

$\tan\theta=\dfrac{b}{a}$

Here, (b) is the opposite side, (a) is the base side.

Pat needs to determine the height of a tree before cutting it down to be sure that it will not fall on a nearby fence.

The angle of elevation of the tree from one position on a flat path from the tree is

$H=40^o$

From a second position,

$I=60\rm\; ft$

Farther along this path it is,

$b=30^o$

Let the distance between tree and first position is x. Thus, from the trigonometry,

$\tan 40=\dfrac{h}{x}\\h=x\tan 40$ ....1

The distance between tree and second position is (x+60) ft. Thus, again from the trigonometry,

$\tan 30=\dfrac{h}{x+60}\\h=(x+60)\tan 40$ ......2

Compare the equation 1 and 2,

$x\tan (40)=(x+60)\tan 30\\0.839x=0.577x+34.64\\0.839x-0.577x=34.64\\x=\dfrac{34.64}{0.262}\\x=132.25\rm\; ft$

Put this value in equation 1 as,

$h=(132.25)\times \tan40\\h\approx111\rm\;ft$

Thus, the height of the tree, which Pat needs to determine, before cutting it down to be sure that it will not fall on a nearby fence is 111 ft.

Learn more about the right angle triangle property here;

brainly.com/question/22790996

#SPJ1

7 0

2 years ago

Which of the following will you need to prepare the claim form

IgorC [24]

<span>C) the explanation of benefits</span>

7 0

3 years ago

complete the system that models the heights of the ball and the receiver’s hands over time. h = -16t² + 14t + h = -16t² + t +

adoni [48]

The height of the ball at time t is given as h = -16t² + 14t + 6 while the height of the receiver hand at time t is given as h = -16t² + 10t + 8

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

A quarterback throws a football toward a receiver from a height of 6 ft. The initial vertical velocity of the ball is 14 ft/s. At the same time that the ball is thrown, the receiver raises his hands to a height of 8 ft and jumps up with an initial vertical velocity of 10 ft/s.

The height of the ball at time t is given as h = -16t² + 14t + 6 while the height of the receiver hand at time t is given as h = -16t² + 10t + 8

Find out more on equation at: brainly.com/question/2972832

#SPJ1

4 0

2 years ago

Antique Accents tracks their daily profits and has found that the distribution of profits is approximately normal with a mean of

mezya [45]

Answer;

a) 0.434

b) 0.983

c) 0.367

Explanation:

The exact question with the given parameters wasn't obtained online, but the same question, albeit with different parameters is then obtained. Hopefully, this Helps to solve the complete question with the required parameters.

Antique Accents tracks their daily profits and has found that the distribution of profis is approximately normal with a mean of $17,700.00 and a standard deviation of about $900.00. Using this information, answer the following questions For full marks your answer should be accurate to at least three decimal places. Compute the probability that tomorrow's profit will be

a) less than $16,791 or greater than $18,231

b) greater than $15,783

c) between $17,997 and $20,130

Solution

This is a normal distribution problem with

Mean = μ = $17,700

Standard deviation = σ = $900

a) less than $16,791 or greater than $18,231. P(x < 16,791) or P(X > 18,231) = P(X < 16,791) + P(x > 18,231)

We first standardize 16,791 and 18,231

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 16791

z = (x - μ)/σ = (16791 - 17700)/900 = - 1.01

For 18231

z = (x - μ)/σ = (18231 - 17700)/900 = 0.59

To determine the required probability

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

We'll use data from the normal probability table for these probabilities

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

P(z < -1.01) = 0.15625

P(z > 0.59) = 1 - (z ≤ 0.59) = 1 - 0.7224 = 0.2776

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59) = 0.15625 + 0.2776 = 0.43385 = 0.434 to 3 d.p

b) greater than $15,783. P(x > 15783)

We standardize 15783

z = (x - μ)/σ = (15783 - 17700)/900 = -2.13

To determine the required probability

P(x > 15783) = P(z > -2.13)

We'll use data from the normal probability table for this probability

P(x > 15783) = P(z > -2.13) = 1 - P(z ≤ - 2.13)

= 1 - 0.01659 = 0.98341 = 0.983 to 3 d.p.

c) between $17,997 and $20,130.

P(17,997 < x < 20,130)

We first standardize 17,997 and 20,130

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 17,997

z = (x - μ)/σ = (17,997 - 17700)/900 = 0.33

For 20,130

z = (x - μ)/σ = (20,130 - 17700)/900 = 2.70

To determine the required probability

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

We'll use data from the normal probability table for these probabilities

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

= P(z < 2.70) - P(z < 0.33)

= 0.99653 - 0.62930

= 0.36723 = 0.367 to 3 d.p.

Hope this Helps!!!

3 0

3 years ago