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NISA [10]
4 years ago
14

Plz help as soon as possible :)

Mathematics
1 answer:
Damm [24]4 years ago
5 0

Answer:

-64

Step-by-step explanation:

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Use rounding (to the nearest 10) to estimate the product of 3217 X 44.
Lorico [155]
So using whole number rounding I believe the answer you should get would be 141,550 as 3217 x 44 is equals to 141,548 so the answer may be 141,550 since 50 is the closest ten number to 48, with a distance of only 2 numbers.
8 0
3 years ago
The selling price of a refrigerator is $693.00. If the markup is 5% of the dealers cost, what is the dealers cost of the refrige
OleMash [197]

Answer:

727.65

Step-by-step explanation: The markup is 5%of the selling price, so multiply the selling price by 0.05 for 5%, you get 34.65. Add that to the selling price, and you get 727.65 as your total.

3 0
3 years ago
11. You pay $24.50 for 10 gallons of gasoline and one
Cloud [144]
Ummm this confuses me a bit
6 0
4 years ago
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]

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

3 0
3 years ago
Solve for the value of X as you should solve for X and explain.
Varvara68 [4.7K]

Answer:

12cos59, about 6.18

Step-by-step explanation:

cos59 = x/12

(adjacent/hypotenuse)

x = 12cos59

plugging this into a calculator gets you about 6.18

6 0
3 years ago
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