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Marina86 [1]
3 years ago
6

David is rowing a boat upstream. The river is flowing at a speed of 2 miles per hour. David starts rowing at a speed of 6 miles

per hour, and his speed decreases at a rate of 1.5 miles per hour.
The equation gives the speed of the boat after David has rowed for x hours? can someone help
y=4+1.5x
y=4-1.5x
y=6-1.5x
Mathematics
2 answers:
jonny [76]3 years ago
8 0
<span>The equation that gives the speed of the boat after David has rowed for x hours is </span><span>y=4-1.5x. 
</span>
The direction David is rowing (upstream) is the positive direction. That makes the direction the river is flowing negative. David rows at 6 mph, but the river is flowing at 2 mph in the opposite direction. That means his "net" speed, relative to an observer on the ground, is 6-2 = 4 mph. He loses 1.5 mph of his speed every hour, so after x hours, his speed is 4-1.5x. 
horsena [70]3 years ago
5 0
Hello there.

<span>David is rowing a boat upstream. The river is flowing at a speed of 2 miles per hour. David starts rowing at a speed of 6 miles per hour, and his speed decreases at a rate of 1.5 miles per hour.
The equation gives the speed of the boat after David has rowed for x hours? can someone help

</span><span>y=4-1.5x</span>
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Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z
dexar [7]

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The value of the constant C is 0.01 .

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Given:

Suppose X, Y, and Z are random variables with the joint density function,

f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0  } \atop {0}; Otherwise} \right.

The value of constant C can be obtained as:

\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1

\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^{-(0.5x + 0.2y + 0.1z)} } \, dz }) \, dy } )\, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y }(\int\limits^\infty_0 {e^{-0.1z} } \, dz  }) \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ])  } \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}]  } \, dy  }) \, dx =1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0  }) \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}]   } \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}]  } \, dx = 1

50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1

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50C[0+\frac{1}{0.5} ] =1

100C = 1 ⇒ C = \frac{1}{100}

C = 0.01

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