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kow [346]
3 years ago
13

Can someone help me with me with my most recent question please

Mathematics
1 answer:
mojhsa [17]3 years ago
8 0

Answer:

no I can't

Step-by-step explanation:

im a dum dum, sorry.

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60 + 7.20 + 7.80 please math is so dumb ​
Alja [10]
75
60+7.20= 67.20
67.20+7.80= 75
6 0
3 years ago
NEED HELP ASAP
Alisiya [41]

Answer:

Divide, 16.

Step-by-step explanation:

So, lets look at the unit scale.

For every 16 ounces, there are 1 pound.

To convert ounces to pounds, the number must be divide by 16, since there are 16 ounces in one pound.

The first fill in box answer is division.

Next we must ifnd the actual amount of pounds.

We know that the box weights 96 ounces.

Now lets divide this by ounces per pound, and see what we get:

96/16

=

6

So the box weighs 6 pounds.

This is the answer to the second fill in question.

Hope this helps!

8 0
3 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
What expression is equivalent to 5y^-3?
sineoko [7]

Answer:

\frac{5}{y^{3}}

Step-by-step explanation:

apply exponent rule:

a^{-b}=\frac{1}{a^b}

7 0
3 years ago
Why must eye protection be worn at all times in welding shop?
kati45 [8]
Because of the ligting of the machine and you can go blind if theres no eye protection
4 0
3 years ago
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