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Hitman42 [59]
2 years ago
15

What is the main advantage of having a skill set with a high market value?

Mathematics
1 answer:
timofeeve [1]2 years ago
3 0

Answer:

In simple words, the main benefit of having a high market value is that one will get a high price for the services he or she will provide. Their job will remain stable and they can lead a good life with the compensation they will receive.

Step-by-step explanation:

thank me later

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Variables x and y are in direct proportion, and y = 12.5 when x = 25. If x = 40, then y =
BabaBlast [244]
25/12.5 = 40/y
cross multiply
(25)(y) = (12.5)(40)
25y = 500
y = 500/25
y = 20 <====
6 0
3 years ago
How is the 4 in 2468 is different from the 4 in 3548
Angelina_Jolie [31]
Because the 4 in “ 2468 “ is in the hundreds place & the 4 in “ 3548” is in the tens place.
3 0
3 years ago
Please help me I am very confused and I can’t get it wrong!!!!
Ann [662]

Answer:

The answer that you are looking for is b

6 0
3 years ago
Use the chain rule calculate dw/dr , dw/ds and dw/dt
Liono4ka [1.6K]

Answer:

\frac{dw}{dr} = \frac{8\cdot r}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}, \frac{dw}{ds} = \frac{10\cdot s}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}, \frac{dw}{dt} = -\frac{2\cdot t}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}

Step-by-step explanation:

We proceed to derive each expression by rule of chain. Let be w = \ln (x+2\cdot y + 3\cdot z), x = r^{2}+t^{2}, y = s^{2}-t^{2} and z = r^{2}+s^{2}:

\frac{dw}{dr} = \frac{\frac{dx}{dr}+2\cdot \frac{dy}{dr} +3\cdot \frac{dz}{dr}  }{x+2\cdot y + 3\cdot z}

\frac{dx}{dr} = 2\cdot r

\frac{dy}{dr} = 0

\frac{dz}{dr} = 2\cdot r

\frac{dw}{dr} = \frac{8\cdot r}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}

\frac{dw}{dr} = \frac{8\cdot r}{4\cdot r^{2}-t^{2}+5\cdot s^{2}} (1)

\frac{dw}{ds} = \frac{\frac{dx}{ds}+2\cdot \frac{dy}{ds} +3\cdot \frac{dz}{ds}  }{x+2\cdot y + 3\cdot z}

\frac{dx}{ds} = 0

\frac{dy}{ds} = 2\cdot s

\frac{dz}{ds} = 2\cdot s

\frac{dw}{ds} = \frac{10\cdot s}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}

\frac{dw}{ds} = \frac{10\cdot s}{4\cdot r^{2}-t^{2}+5\cdot s^{2}} (2)

\frac{dw}{dt} = \frac{\frac{dx}{dt}+2\cdot \frac{dy}{dt} +3\cdot \frac{dz}{dt}  }{x+2\cdot y + 3\cdot z}

\frac{dx}{dt} = 2\cdot t

\frac{dy}{dt} = -2\cdot t

\frac{dz}{dt} = 0

\frac{dw}{dt} = -\frac{2\cdot t}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}

\frac{dw}{dt} = -\frac{2\cdot t}{4\cdot r^{2}-t^{2}+5\cdot s^{2}} (3)

7 0
2 years ago
For a standard normal distribution find the approximate value of p(z&lt;0.42)
Kruka [31]

Answer:

p(z<0.42) = 0.6628

Step-by-step explanation:

A normal distribution with a mean of 0 and a standard deviation of 1 is called a standard normal distribution. That is to say:

μ = 0

σ² = 1

Using a calculator, we find that:

p(z<0.42) = 0.6628 (See picture attached)

8 0
3 years ago
Read 2 more answers
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