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12345 [234]
1 year ago
14

12 A pilot needs to know the maximum height an aircraft can fly at. The cabin has been tested and is safe to a height of 15679 m

. Round this height appropriately: a) to a whole number of kilometres b) to a whole 100 m.please help me​
Mathematics
1 answer:
abruzzese [7]1 year ago
7 0
15.6 kilometers rounded down because it is not safe to go above what has been approved
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Tammy accidentally wrote a check for more money than she had in her checking account. When she got her statement, she saw that h
sergejj [24]

Answer:

-100

Step-by-step explanation:

1/5 of -125 will be -25.

So, if she deposits 1/5 of what she owes, then she will deposit $25.

Her balance will then become -100 because she paid back 25.

Her new balance is -100

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3 years ago
I NEED HELP ILL GIVE BRAINLEST
IRISSAK [1]
The last 2 options are correct
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The capacity of an elevator is 12 people or 2088 pounds. the capacity will be exceeded if 12 people have weights with a mean gre
boyakko [2]

suppose the people have weights that are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.

Find the probability that if a person is randomly selected, his weight will be greater than 174 pounds?

Assume that weights of people are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.

Mean = 177

standard deviation = 26

We find z-score using given mean and standard deviation

z = \frac{x-mean}{standard deviation}

= \frac{174-177}{26}

=-0.11538

Probability (z>-0.11538) = 1 - 0.4562 (use normal distribution table)

= 0.5438

P(weight will be greater than 174 lb) = 0.5438

8 0
3 years ago
What is the momentum of 200kg portable if it travels 50 meters in 5seconds
kvasek [131]
Momentum  = mass * velocity

50 meters in 5 seconds = 50/5  = 10 m / second

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5 0
3 years ago
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
3 years ago
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