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3 years ago

Help me guys........m

Mathematics

2 answers:

zzz [600]3 years ago

7 0

They are g and a hope this helped

Gekata [30.6K]3 years ago

7 0

The answer for the middle question is H I did it on a graphing calculator

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Rooms in a hotel are numbered from 1 to 19.

skelet666 [1.2K]

Answer:

a) 8/19 chance

b) 7/19 chance (I'm not 100% sure)

Step-by-step explanation:

A) since there are 8 prime numbers from 1-19 we can say that the probability is 8 over 19 or 8/19

4 0

2 years ago

A 2-digit number is one more than 6 times the sum of its digits. If the digits are reversed, the new number is 9 less than the o

kramer

Answer:

I think the answer is 43

8 0

2 years ago

How many lines of symmetry?

Alex73 [517]

Answer:

i think 4

sorryif its wrong

3 0

2 years ago

Read 2 more answers

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

Suppose that 1 kg of Sodium-24 is reduced to 0.0156 kg in 90 hours. What is the half-life, in hours, of Sodium-24?

N76 [4]

Answer:

14.99 hours

Step-by-step explanation:

The formula for half-life is given as,

R/R' = 2ᵃ/ᵇ...................... Equation 1

Where R= Original mass of sodium-24, R' = mass of Sodium-24 after disintegration, a = Total time taken for sodium-24 to disintegrate, b = half-life of sodiun-24.

Given: R = 1 kg, R' = 0.0156 kg, a = 90 hours

Substitute into equation 1 and solve for b

1/(0.0156) = 2⁹⁰/ᵇ

Taking the log of both side

log[1/(0.0156)] = log(2⁹⁰/ᵇ)

log[1/(0.0156)] = 90/b(log2)

90/b = log[1/(0.0156)]/log2

90/b = 6.0023

b = 90/6.0023

b = 14.99 hours

b = 14.99 hours.

Hence the half-life of Sodium-24 is 14.99 hours

7 0

2 years ago