The mean of a normally distributed dataset is 12, and the standard deviation is 2.what ? % of the data points lies between 8 and 16.

Answers

Answer 1

Answer: In the Normal distribution: if the mean (M) is 12 and the standard deviation (s)is 2, then between 8 (12- 2*2, or M - 2 s) and 16 (12 + 2*2, or M+2 s ),or between M - 2 s and M + 2 s lies 13.6 %+ 34.1% + 34.1 % + 13.6% = 95.4% data points. Graph is in the attachment.

Answer 2

Answer:

The correct answer is:

95%

Explanation:

The empirical rule states that 68% of normally distributed data lie within 1 standard deviation of the mean; 95% lie within 2 standard deviations of the mean; and 99.7% lie within 3 standard deviations of the mean.

The mean is 12 and the standard deviation is 2. This means that values from 10 to 14 (12-2, 12+2) are within 1 standard deviation of the mean.

Values from 8 to 16 (12-2-2, 12+2+2) are within 2 standard deviations of the mean. These are the values in question; this means that 95% of the data lie within this range.

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The one to one functions G and H are defined as followG={(-8,1),(1,5),(4,-5)(8,-9)}
H(x)=3x+4
find the following
g^-1(1)=
h^-1(x)=
(h^-1oh)(-2)=

Answers

$G=\{(-8,1),(1,5),(4,-5),(8,-9)\}\n\n\Rightarrow\ \ \ G^(-1)=\{(1,-8),(5,1),(-5,4),(-9,8)\}\ \ \ \Rightarrow\ \ \ g^(-1)(1)=-8\n\nH(x)=3x+4\ \ \ and\ \ \ D_H=R\n\n 3x=H(x)-4\ \ \ \Rightarrow\ \ \ x= (H(x)-4)/(3)\n\nh^(-1)(x)= (x-4)/(3) \ \ \ and\ \ \ D_(h^(-1))=R\n\n\n(h^(-1)\circ h)(-2)=-2\n\nh^(-1)\bigg {(}h(-2)\bigg{)}=h^(-1)\bigg {(}3(-2)+4\bigg{)}=h^(-1)(-2)= (-2-4)/(3) = (-6)/(3) =-2$

If triangle STU is similar to triangle VWX, which statement is true about the two triangles?A. Segment ST is congruent to segment VW, and angles U and X are proportional.

B. Segment TU is proportional to segment WX, and angles S and V are proportional

C. segment ST is congruent to segment VW, and angles U and X are congruent.

D. segment TU is proportional to segment WX, and angles S and V are congruent.

Answers

B. Segment TU is proportional to segment WX, and angles S and V are proportional

Answer:B. Segment TU is proportional to segment WX, and angles S and V are proportional

Step-by-step explanation:

Can anyone explain how to differentiate this ? (2x^2+3)sin5x

Answers

Hello,

(u*v)'=u'v+uv'
u=2x²+3 ==> u'=4x
v=sin 5x ==>v'=cos 5x *5

((2x²+3 sin (5x))'=4x* sin (5x) +(2x²+3)*cos (5x) * 5

I need help with these could you guys help me?

Answers

The first one simplifies by collecting like terms to become:
7a²(4b²+9ab+10)
and the second one simplifies again by collecting like terms to become:
7u(16u²-49v²)
Got anymore questions, happy to help :)

Which set of numbers is included as part of the solution set of the compound inequality x < 6 or x > 10?
{–7, –1.7, 6.1, 10}
{–3, 4.5, 13.6, 19}
{0, 6, 9.8, 14}
{8.5, 9.1}

Answers

All of the numbers in the set must either be less than 6 or greater than 10. They can't be between 6 and 10. The only set that works is {-3, 4.5, 13.6, 19}

Answer:

its option b ( -3,4.5,13.6,19)

Step-by-step explanation:

just did it on edge 2021 <3

If the work required to stretch a spring 2 ft beyond its natural length is 12 ft-lb, how much work is needed to stretch it 9 in. beyond its natural length?

Answers

The amount of work required to stretch 9 inches beyond the natural length will be 4.5 ft-lb

Given data:

To determine the work required to stretch the spring 9 inches beyond its natural length, use the concept of Hooke's Law.

Hooke's Law states that the force required to stretch or compress a spring is directly proportional to the displacement from its equilibrium position.

Given that stretching the spring by 2 ft requires 12 ft-lb of work, determine the constant of proportionality.

The constant of proportionality (k) represents the stiffness of the spring and can be calculated using the formula:

k = work / displacement

k = 12 ft-lb / 2 ft

k = 6 lb/ft

Now, calculate the work required to stretch the spring 9 inches (0.75 ft) beyond its natural length using the same constant of proportionality:

work = k * displacement

work = 6 lb/ft * 0.75 ft

work = 4.5 ft-lb

Hence, it would require 4.5 ft-lb of work to stretch the spring 9 inches beyond its natural length.

To learn more about Hooke's law, refer:

brainly.com/question/30379950

#SPJ12

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