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Irina18 [472]
3 years ago
8

Suppose each runner ran at the rate given in the table above for 3.1 miles. How much time will elapse between the first place fi

nisher and the last place finisher?
Mathematics
1 answer:
Scilla [17]3 years ago
6 0
You didn't include the table but I found this table for the same statement, so I will answer you based on the next table:

Runner         distance           time

Arabella        7,299 feet        561 seconds
Bettina          3,425 yards     13 minutes, 12 seconds
Chandra       8,214 feet        0,195 hours
Divya            1,62 miles        732 seconds

To answer the question you must find the rate for each runner and then calculate the time to run 3.1 miles at each rate.

First you need to convert the data to obtain the rate in miles per second.

These are the main conversion identities:

1 mile = 5280 feet

1 mile = 1760 yards

1 hour = 3600 seconds

1 hour = 60 minutes

1 minute = 60 seconds

Arabella:

rate: 7,229 feet / 561 seconds * (1 mile / 5280 feet)  =

= 0.00244 mile/second

Time to run 3.1 miles: V = d / t => t = d / V = 3.1 miles / 0.00244 mile/second = 1270 seconds

Bettina:

13 minutes + 12 seconds = 13*60 seconds +12 seconds = 792 seconds

rate = 3425 yards / 792 seconds * 1 mile / 1760 yards = 0.00246 mile/seconds

Time to run 3.1 miles = 3.1 miles / 0.00246 mile/second = 1260 seconds

Chandra:

rate = 8214 feet / 0.195 hours * 1 mile / 5280 feet * 1hour / 3600 seconds =

= 0.00222 seconds

Time = 3.1 mile / 0.00222 seconds = 0.389 hour = 1396 seconds

Divya:

rate = 1.62 miles / 732 seconds = 0.00221 seconds

Time = 3.1 mile / 0.00221 seconds = 1403 seconds

Now you can find the difference between fhe last and the first 1403 seconds - 1260 seconds = 143 seconds

That is equivalent to 2.38 seconds.
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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
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Answer:

The value of the constant C is 0.01 .

Step-by-step explanation:

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

50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1

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

100C = 1 ⇒ C = \frac{1}{100}

C = 0.01

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