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icang [17]
2 years ago
5

Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. 27% of the possible Z values are grea

ter than _____________.
Mathematics
1 answer:
inessss [21]2 years ago
6 0

Answer:

27% of the possible Z values are greater than 0.613

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 0, \sigma = 1

27% of the possible Z values are greater than

The 100 - 27 = 73rd percentile, which is X when Z has a pvalue of 0.73. So X when the z-score is 0.613.

Z = \frac{X - \mu}{\sigma}

0.613 = \frac{X - 0}{1}

X = 0.613

27% of the possible Z values are greater than 0.613

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marshall27 [118]
Any number times 0 is going to equal 0
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2 years ago
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Find the dimensions of a rectangle with area 512 m2 whose perimeter is as small as possible. (If both values are the same number
Masja [62]

Answer:

<h2>√512 by √512 </h2>

Step-by-step explanation:

Length the length and breadth of the rectangle be x and y.

Area of the rectangle A = Length * breadth

Perimeter P = 2(Length + Breadth)

A = xy and P = 2(x+y)

If the area of the rectangle is 512m², then 512 = xy

x = 512/y

Substituting x = 512/y into the formula for calculating the perimeter;

P = 2(512/y + y)

P = 1024/y + 2y

To get the value of y, we will set dP/dy to zero and solve.

dP/dy = -1024y⁻² + 2

-1024y⁻² + 2 = 0

-1024y⁻² = -2

512y⁻² = 1

y⁻² = 1/512

1/y² = 1/512

y²  = 512

y = √512 m

On testing for minimum, we must know that the perimeter is at the minimum when y = √512

From xy = 512

x(√512) = 512

x = 512/√512

On rationalizing, x = 512/√512 * √512 /√512

x = 512√512 /512

x = √512 m

Hence, the dimensions of a rectangle is √512 m  by √512 m

5 0
2 years ago
Caleb has 20 more books than Danica. If Caleb gives Danica half of his books, Danica will have 4 more books than Caleb. How many
wariber [46]

Answer:

Option E: 24.

Step-by-step explanation:

When Caleb (C) has 20 more books than Danica (D), we have:

C = 20 + D   (1)

Now, when Caleb gives Danica half of his books:

C' = C - \frac{C}{2}   (2)

Danica will have 4 more books than Caleb, so:

D' = D + \frac{C}{2}     (3)    

Also we have that:

D' = C' + 4   (4)  

By entering equation (2) and equation (3) into equation (4) we have:

D + \frac{C}{2} = C - \frac{C}{2} + 4   (5)

From (1) we have:

D = C - 20   (6)

By entering (6) into (5):

C - 20 + \frac{C}{2} = C - \frac{C}{2} + 4

C = 24

And, from (6):

D = C - 20 = 24 - 20 = 4  

Therefore, the correct option is E, Caleb had to start with 24 books.

I hope it helps you!          

3 0
2 years ago
Graph a system of equations to solve log (−5.6x + 1.3) = −1 − x. Round to the nearest tenth. From the least to the greatest, the
natima [27]

Answer:

  • See the graph attached
  • x₁ ≈ - 2.1
  • x₂ ≈ 0.2

Explanation:

To solve log (−5.6x + 1.3) = −1 − x graphycally, you must graph this system of equations on the same coordinate plane:

  • Equation 1: y = log (5.6x + 1.3)
  • Equatin2:    y = - 1 - x

1) To graph the equation 1 you can use these features of logarithmfunctions:

  • Domain: positive values ⇒ -5.6x + 1.3 > 0 ⇒ x < 13/56 (≈ 0.23)

  • Range: all real numbers (- ∞ , ∞)

  • x-intercept:

        log ( -5.6x + 1.3) = 0 ⇒ -5.6x + 1.3 = 1 ⇒x = 0.3/5.6 ≈ 0.054

  • y-intercept:

       x = 0 ⇒ log (0 + 1.3) = log (1.3) ≈ 0.11

  • Pick some other values and build a table:

        x            log (-5.6x + 1.3)

        -1           0.8

        -2           1.1

        -3           1.3

  • You can see such graph on the picture attached: it is the red curve.

2) Graphing the equation 2 is easier because it is a line: y = - 1 - x

  • slope, m = - 1 (the coeficient of x)
  • y - intercept, b = - 1 (the constant term)
  • x - intercept: y = 0 = - 1 - x ⇒ x = - 1
  • The graph is the blue line on the picture.

3) The solution or solutions of the equations are the intersection points of the two graphs. So, now the graph method just requires that you read the x coordinates of the intersection points. From the least to the greatest, rounded to the nearest tenth, they are:

  • <u><em>x₁ ≈ - 2.1</em></u>
  • <u><em>x₂ ≈ 0.2</em></u>

8 0
2 years ago
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Show that the given set of functions is orthogonal with respect to the given weight on the prescribed interval. Find the norm of
AnnyKZ [126]

Answer/Explanation

The complete question is:

Show that the set function {1, cos x, cos 2x, . . .} is orthogonal with respect to given weight on the prescribed interval [- π, π]

Step-by-step explanation:

If we make the identification For ∅° (x) = 1 and  ∅n(x) = cos nx, we must show that ∫ lim(π) lim(-π) .∅°(x)dx = 0 , n ≠0, and ∫ lim(π) lim(-π) .∅°(x)dx = 0, m≠n.

Therefore, in the first case, we have

(∅(x), ∅(n)) ∫ lim(π) lim(-π) .∅°(x)dx = ∫ lim(π) lim(-π) cosn(x)dx

This will therefore be equal to :

1/n sin nx lim(π) lim(-π) = 1/n  [sin nπ - sin(-nπ)] = 0 , n ≠0 (In the first case)

and in the second case, we have,,

(∅(m) , ∅(n)) = ∫ lim(π) lim(-π) .∅°(x)dx

This will therefore be equal to:

∫ lim(π) lim(-π) cos mx cos nx dx

Therefore, 1/2 ∫ lim(π) lim(-π)( cos (m+n)x + cos( m-n)x dx (Where this equation represents the trigonometric function)

1/2 [ sin (m+n)x / m+n) ]+ [ sin (m-n)x / m-n) ]  lim(π) lim(-π) = 0, m ≠ n

Now, to go ahead to find the norms in the given set intervals, we have,

for  ∅°(x) = 1 we have:

//∅°(x)//² = ∫lim(π) lim(-π) dx = 2π

So therefore, //∅°(x)//² = √2π

For ∅°∨n(x)  = cos nx  , n > 0.

It then follows that,

//∅°(x)//² = ∫lim(π) lim(-π) cos²nxdx = 1/2 ∫lim(π) lim(-π) [1 + cos2nx]dx = π

Thus, for n > 0 , //∅°(x)// = √π

It is therefore ggod to note that,

Any orthogonal set of non zero functions {∅∨n(x)}, n = 0, 1, 2, . . . can be  normalized—that is, made into an orthonormal set by dividing each function by  its norm. It follows from the above equations that has been set.

Therefore,

{ 1/√2π , cosx/√π , cos2x/√π...} is orthonormal on the interval {-π, π}.

6 0
2 years ago
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