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2 years ago

Sin30=4/sqrt{3}/x How to solve?

Mathematics

1 answer:

iragen [17]2 years ago

6 0

Answer:

$x=\dfrac{8}{\sqrt3}$

Step-by-step explanation:

Given that,

$\sin(30)=\dfrac{\dfrac{4}{\sqrt3}}{x}$ ...(1)

We need to find the value of x.

We know that,

sin30 = 1/2

Use it in equation (1).

$\dfrac{1}{2}=\dfrac{\dfrac{4}{\sqrt3}}{x}\\\\x=\dfrac{4}{\sqrt3}\times 2\\\\x=\dfrac{8}{\sqrt3}$

So, the value of x is equal to $\dfrac{8}{\sqrt3}$ .

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ser-zykov [4K]

IT IS OPTION B

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2 years ago

Read 2 more answers

John is 36 years old now. 1/4 of his age is equal to 1/3 of sam's age. find their total age in 5 years time

sveticcg [70]

Answer:

John = 41 years

Sam = 32 years

Step-by-step explanation:

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2 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

2 years ago

Vending machines can be adjusted to reject coins above and below certain weights. The weights of legal U.S. quarters have a norm

tigry1 [53]

Answer: The percentage of legal quarters will be rejected by the vending machine = 2.275%

Step-by-step explanation:

Given: The weights of legal U.S. quarters have a normal distribution with a mean of 5.67 grams and a standard deviation of 0.07 gram.

Let x be the weights of legal U.S. quarters .

Required probability: $P(x5.81)$

$=P(\dfrac{x-\mu}{\sigma}\dfrac{5.81-5.67}{0.07})\\\\=P(z2)\approx0.02275 \ \ \ [By\ P-value\ calculator]$

The percentage of legal quarters will be rejected by the vending machine = 2.275%

5 0

2 years ago

What is sin 0 when sec 0 = 5? 0 |(1,0) sin e = 2[?] [] Rationalize the denominator if necessary.

Komok [63]

Answer:

$Sin ~0 =\frac{2\sqrt{5} }{5}$

Step-by-step explanation:

$sec0=\sqrt{5}$

$cos~0=\frac{1}{sec0} =\frac{1}{\sqrt{5} }$

$sin ~0=\sqrt{1-cos^2~0}$

$=\sqrt{1-1/5}=\sqrt{5-1/5}$

$Sin~0=\sqrt{4/5} =2/\sqrt{5}*\sqrt{5}/\sqrt{5}$

$Sin~0=2\sqrt{5}/2=2\sqrt{5}/5$

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hope it helps...

have a great day!!!

7 0

3 years ago