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anzhelika [568]
3 years ago
7

Assume that women’s heights are normally distributed with a mean given by μ=63.5

Mathematics
2 answers:
slavikrds [6]3 years ago
5 0

Answer:64.3738

Step-by-step explanation:

xz_007 [3.2K]3 years ago
3 0

Answer:

Part a) 0.5636  or 56.36%

Step-by-step explanation:

Part a)

Given the info on mean and standard deviation, we evaluate the Z-score as:

Z = (64 - 63.5)/3.1 = 0.16

and use the Z-tables to find that

P(x < 64) = 0.5636  = 56.36%

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Can somebody explain how these would be done? The selected answer is incorrect, and I was told "Nice try...express the product b
trapecia [35]

Answer:

Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) \cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}

( 2 ) \sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}

These two identities makes sin(π / 10) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and cos(π / 10) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}.

Therefore cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}. Substitute,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right]

And now simplify this expression to receive our answer,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right] = -\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i,

-\frac{3\sqrt{5+\sqrt{5}}}{4} = -2.01749\dots and \:\frac{3\sqrt{3-\sqrt{5}}}{4} = 0.65552\dots

= -2.01749+0.65552i

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

First Attachment : We know from the previous problem that cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}, cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,

6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}

We know that 6\sqrt{5+\sqrt{5}} = 16.13996\dots and -\:6\sqrt{3-\sqrt{5}} = -5.24419\dots . Therefore,

Solution : 16.13996 - 5.24419i

Which rounds to about option b.

7 0
3 years ago
Type the correct answer in the box. use numerals instead of words. if necessary, use / for the fraction bar. find the missing te
Allisa [31]

The missing term in the provided quadratic equation is 10x if the roots of a quadratic equation are 5 ± 3i.

<h3>What is a complex number?</h3>

It is defined as the number which can be written as x+iy where x is the real number or real part of the complex number and y is the imaginary part of the complex number and i is the iota which is nothing but a square root of -1.

The question is incomplete.

The complete question is in the picture, please refer to the attached picture.

We have the roots of a quadratic equation:

5 ± 3i

To find the quadratic equation:

(x - (5+3i))(x - (5-3i))

\rm =x^2+x\left(-\left(5-3i\right)\right)-\left(5+3i\right)x-\left(5+3i\right)\left(-\left(5-3i\right)\right)

= x² -10x + 34

The missing value is 10x

The quadratic equation is:

= x² -10x + 34

Thus, the missing term in the provided quadratic equation is 10x if the roots of a quadratic equation are 5 ± 3i.

Learn more about the complex number here:

brainly.com/question/10251853

#SPJ1

6 0
2 years ago
Julio believes that the quotient of 6 ÷ 1/2 = 3
Hitman42 [59]

Answer:

Julio did 6/2. The correct answer is 12.

Step-by-step explanation:

6/1/2=6*2=12. Julio forgot to change the division into multiplication. Dividing is multiplying by it's reciprocal.

5 0
2 years ago
Read 2 more answers
6. Which value is a solution to the inequality? −3x −5 &lt; −14
yan [13]
First, add -5 to -14: 

-3x < -9 

Then, just divide!: 

x < 3
Hopefully that helped! :) 
8 0
3 years ago
Read 2 more answers
An underground conical tank has a top diameter of 4m and a depth of 6m. Compute the capacity of the tank in cubic meter.
dlinn [17]

Step-by-step explanation:

1/3πr^2h

{(1/3)×(22/7)×2^2×6}m^3

[(1/3)×(22/7)×4×6)}m^3

{528/21}m^3

=25.1 m^3

NOTE:r=diameter divided by 2

7 0
3 years ago
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