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netineya [11]
2 years ago
10

How many groups of 1/2 are in 3 1/2

Mathematics
1 answer:
Nuetrik [128]2 years ago
7 0

Answer:

7 Groups of 1/2 in 3 1/2

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Please guys help me out . Thanks
Naya [18.7K]

Answer:

6.7

Step-by-step explanation:

cos(48) = adjacent/hypotenuse

cos(48) = x/10

x = 10cos(48)

x = 6.691306064

6 0
3 years ago
Gwen measured a kitchen and made a scale drawing. The scale she used was 11 inches : 5 feet. If a countertop in the kitchen is 2
aliina [53]

Answer:

11x 5 =55-22=33

Step by step explanation

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8 0
2 years ago
What is the midpoint of the segment with two endpoints (4,-6) and (-2,-3)? Put your answers in decimal from, if applicable.
Ber [7]
Answer: 1,-4.5

Explanation: you can find this by finding the difference in y and the difference in x. The difference in x is -6, and the difference in y is +3. All you have to do after that is divide that by 2: -3,1.5, than add that to the first coordinate. That should be the midpoint.
4 0
1 year 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
2 years ago
The table below shows the number of school absences each day.
Vladimir79 [104]

Answer

A

Step-by-step explanation:

Did it on edge20202

4 0
2 years ago
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