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3 years ago

1. Find the distance between each pair of points. Round your answer to

Mathematics

1 answer:

zlopas [31]3 years ago

7 0

Step-by-step explanation:

$\sqrt{(3 - ( - 4))^{2} + ( - 7 - 6)^{2} } \\ = \sqrt{(3 + 4)^{2} + {( - 13)}^{2} } \\ = \sqrt{ 49 + 169} \\ = \sqrt{218} \\ = 14.7648$

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What's the ratio 48:30

Vladimir79 [104]

24:15 would be the simplified version, if that's what you meant.

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3 years ago

Find the 12th term of an AP of the form 7, 12, 17, 22

ikadub [295]

Answer: 62

Step-by-Step Explanation:

First Term (a) = 7
Common Difference (d) = 12 - 7 = 5
Term to Find (n) = 12th

Therefore, finding the 12th Term :-
=> a+(n-1)d
= 7 + (12 - 1)5
= 7 + (11)5
= 7 + 55
=> 62

Hence, 12th Term of this AP is 62

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2 years ago

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Cho tam giác ABC vuông tại C, biết : Góc A = 50 độ. Tính góc B

chubhunter [2.5K]

Answer:

Tổng 3 góc trong 1 tam giác là 180 độ

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2 years ago

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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]

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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3 years ago

What is the probabibility of rolling a sum of 10 when rolling two number cubes?

Alja [10]

Answer: 1 over 12

Step-by-step explanation:

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3 years ago

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