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grandymaker [24]
2 years ago
10

Fiona finds that the distance between two cities is 83,400,000 millimeters. What is the distance between the two cities in kilom

eters? Use the metric table to help answer the question.
Mathematics
2 answers:
daser333 [38]2 years ago
8 0
The metric table goes millimeters, centimeters, decimeters, meters, dekameters, hectometers, kilometers. To move up the table, move the decimal point one to the left. If you wanted to change it to centimeters, the answer would be 8,340,000. In kilometers, this is 83.4 kilometers. Hope this helps!
yawa3891 [41]2 years ago
4 0

Answer:

83.4 kilometers. Hope this help!!!!

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Jonah is thinking of a 2-digit number. It is a multiple of 6 and 12. It is a factor of 108. The sum of its digits is 9. What num
Colt1911 [192]

Answer:

Step-by-step explanation:

36

36x3=108

3+6=9

6 0
2 years ago
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19x+rx=-37+w <br><br><br> solve for x
ipn [44]

Factor the left side:-

x (19 + r) = -37 + w

x = ( -37 + w) / (19 + r) Answer


3 0
2 years ago
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A rectangular prism has a volume of 170 cubic centimeters. The length of the prism is 5 centimeters, and the height of the prism
luda_lava [24]

Answer:

The width of the prism is 2 cm

Step-by-step explanation:

The given parameters are;

The volume of the prism = 170 cm³

The length of the prism = 5 cm

The height of the prism = 17 cm

The volume of the prism is given by the  relationship v = Length, l × Height, h × Width, w

Therefore;

The volume of the prism = 5 cm × 17 cm × w = 170 cm³

Which gives;

w = 170 cm³/(5 cm × 17 cm) = 170 cm³/(85 cm) = 2 cm

∴ The width of the prism = 2 cm.

7 0
2 years ago
Assume that the ages for first marriages are normally distributed with a mean of 26 years and a standard deviation of 4 years. W
Snowcat [4.5K]

Answer:

0.7743

Step-by-step explanation:

Mean of age = u = 26 years

Standard Deviation = \sigma = 4 years

We need to find the probability that the person getting married is in his or her twenties. This means the age of the person should be between 20 and 30. So, we are to find P( 20 < x < 30), where represents the distribution of age.

Since the data is normally distributed we can use the z distribution to solve this problem. The formula to calculate the z score is:

z=\frac{x-u}{\sigma}

20 converted to z score will be:

z=\frac{20-26}{4}=-1.5

30 converted to z score will be:

z=\frac{30-26}{4}=1

So, now we have to find the probability that the z value lies between -1.5 and 1.

P( 20 < x < 30) = P( -1.5 < z < 1)

P( -1.5 < z < 1 ) = P(z < 1) - P(z<-1.5)

From the z-table:

P(z < 1) = 0.8413

P(z < -1.5) =0.067

So,

P( -1.5 < z < 1 ) = 0.8413 - 0.067 = 0.7743

Thus,

P( 20 < x < 30) = 0.7743

So, we can conclude that the probability that a person getting married for the first time is in his or her twenties is 0.7743

6 0
2 years ago
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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