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2 years ago

Write 103727495 in word form

Mathematics

2 answers:

OLga [1]2 years ago

8 0

One hundred and three million, seven hundred and twenty-seven thousand, four hundred and ninety five.

ss7ja [257]2 years ago

8 0

One hundred and three million, seven hundred and twenty seven thousand, four hundred and ninety five.

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Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane

liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$T(x,y) =x\sin2y$

$r = 1m$

$v = 2m/s$

Express the given point P as a unit tangent vector:

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$u = \frac{\sqrt 3}{2}i - \frac{1}{2}j$

Next, find the gradient of P and T using: $\triangle T = \nabla T * u$

Where

$\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j$

So: the gradient becomes:

$\triangle T = \nabla T * u$

$\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]$

By vector multiplication, we have:

$\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}$

$\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)$

$\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5$

$\triangle T = 0.935042$

Hence, the rate is:

$Rate = \triangle T = 0.935042^\circ /cm$

3 0

2 years ago

Which exponential function goes through the points (1, 8) and (4, 64)?

Mama L [17]

An exponential function can be expressed as:

y=ab^x we have two points so we can solve

64/8=(ab^4)/(ab^1)

8=b^3

2=b, now we have:

y=a(2^x), using point (4,64)

64=a(2^4)

64=16a

4=a so

y=4(2^x)

6 0

2 years ago

Easy but i'm confused yikes. (picture attached)

kirill115 [55]

Answer:

The second answer

Step-by-step explanation:

P'Q'R' is simply rotating PQR 90 degrees, it does not change the angle that it makes

8 0

3 years ago

Read 2 more answers

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

Marrrta [24]

Answer:

a) Bi [P ( X >=15 ) ] ≈ 0.9944

b) Bi [P ( X >=30 ) ] ≈ 0.3182

c) Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) Bi [P ( X >40 ) ] ≈ 0.0046

Step-by-step explanation:

Given:

- Total sample size n = 745

- The probability of success p = 0.037

- The probability of failure q = 0.963

Find:

a. 15 or more will live beyond their 90th birthday

b. 30 or more will live beyond their 90th birthday

c. between 25 and 35 will live beyond their 90th birthday

d. more than 40 will live beyond their 90th birthday

Solution:

- The condition for normal approximation to binomial distribution:

n*p = 745*0.037 = 27.565 > 5

n*q = 745*0.963 = 717.435 > 5

Normal Approximation is valid.

a) P ( X >= 15 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( X >=15 ) ] = N [ P ( X >= 14.5 ) ]

- Then the parameters u mean and σ standard deviation for normal distribution are:

u = n*p = 27.565

σ = sqrt ( n*p*q ) = sqrt ( 745*0.037*0.963 ) = 5.1522

- The random variable has approximated normal distribution as follows:

X~N ( 27.565 , 5.1522^2 )

- Now compute the Z - value for the corrected limit:

N [ P ( X >= 14.5 ) ] = P ( Z >= (14.5 - 27.565) / 5.1522 )

N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 )

- Now use the Z-score table to evaluate the probability:

P ( Z >= -2.5358 ) = 0.9944

N [ P ( X >= 14.5 ) ] = P ( Z >= -2.5358 ) = 0.9944

Hence,

Bi [P ( X >=15 ) ] ≈ 0.9944

b) P ( X >= 30 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( X >=30 ) ] = N [ P ( X >= 29.5 ) ]

- Now compute the Z - value for the corrected limit:

N [ P ( X >= 29.5 ) ] = P ( Z >= (29.5 - 27.565) / 5.1522 )

N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 )

- Now use the Z-score table to evaluate the probability:

P ( Z >= 0.37556 ) = 0.3182

N [ P ( X >= 29.5 ) ] = P ( Z >= 0.37556 ) = 0.3182

Hence,

Bi [P ( X >=30 ) ] ≈ 0.3182

c) P ( 25=< X =< 35 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( 25=< X =< 35 ) ] = N [ P ( 24.5=< X =< 35.5 ) ]

- Now compute the Z - value for the corrected limit:

N [ P ( 24.5=< X =< 35.5 ) ]= P ( (24.5 - 27.565) / 5.1522 =<Z =< (35.5 - 27.565) / 5.1522 )

N [ P ( 24.5=< X =< 25.5 ) ] = P ( -0.59489 =<Z =< 1.54011 )

- Now use the Z-score table to evaluate the probability:

P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

N [ P ( 24.5=< X =< 35.5 ) ]= P ( -0.59489 =<Z =< 1.54011 ) = 0.6623

Hence,

Bi [P ( 25=< X =< 35 ) ] ≈ 0.6623

d) P ( X > 40 ) ?

- Apply continuity correction for normal approximation:

Bi [P ( X >40 ) ] = N [ P ( X > 41 ) ]

- Now compute the Z - value for the corrected limit:

N [ P ( X > 41 ) ] = P ( Z > (41 - 27.565) / 5.1522 )

N [ P ( X > 41 ) ] = P ( Z > 2.60762 )

- Now use the Z-score table to evaluate the probability:

P ( Z > 2.60762 ) = 0.0046

N [ P ( X > 41 ) ] = P ( Z > 2.60762 ) = 0.0046

Hence,

Bi [P ( X >40 ) ] ≈ 0.0046

4 0

3 years ago

A line passes through the points (-4,5) and (1,-5). Write the equation of the line in slope-

sweet [91]

Answer:

y = -2x - 3

Step-by-step explanation:

Slope: (5 - - 5)/(-4 -1) = 10/-5 = -2

y-intercept: 5 - (-2)(-4) = -3

4 0

2 years ago