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3 years ago

Which exponential equation is equivalent to the logarithmic equation below? log 478= a

1 answer:

5 0

$\hbox{if } \log_x y=z \hbox{ then } x^z=y \\ \\
\log 478=a \\ \downarrow \\
\log_{10} 478=a \\ \Downarrow \\ \boxed{10^a=478}$

The answer is A.

You might be interested in

Sarah biked the same distance each day for 6 days. If she traveled 468 kilometers altogether, how many kilometers did she travel

I am Lyosha [343]

76 kilometers per day.

Step by step

468 divided by 6 = 76

76•6=468

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

jill traveled to the ferry office and back. the trip there took 5 hours and the trip back took 4 hours. she averaged 65 mph on t

antoniya [11.8K]

We know from physics class that the formula for distance of a linear motion is given as:

d = v t

Where,

d = distance travelled

v = average velocity

t = time it took to reach the destination

Since the distance going to the office and back is just similar, therefore we can simply equate the two:

v1 t1 = v2 t2

Where 1 signifies going to the office and 2 signifies going back from the office. Therefore this yields to:

v1 * 5 hours = 65 mph * 4 hours

v1 = 52 mph

Answer: The average speed going to the office is 52 mph.

4 0

3 years ago

Please help me out!!!!

Serggg [28]

Answer:

If you are trying to solve by substitution then you have 2 different possibilities of answer forms.

First Possibility: (Point form) (16,8)

Second Possibility: (equation form) x=16, y=8

PLEASE MARK BRAINLIEST

7 0

3 years ago

uppose a large telephone manufacturer has a problem with excessive customer complaints and consequent returns of the phones for

vodka [1.7K]

Answer:

The answer is "294.2075".

Step-by-step explanation:

Given:

$\hat{p} = 0.5 \\\\1 - \hat{p} = 1 - 0.5 = 0.5\\\\E = 4\% = 0.04\\\\$

At $83\%$ confidence level the z is ,

$\alpha = 1 - 83\% = 1 - 0.83 = 0.17\\\\\frac{\alpha}{2} = \frac{0.17}{2} = 0.085\\\\Z_{\frac{\alpha}{2}} = Z_{0.085} = 1.3722\\\\n = (\frac{Z_{\frac{\alpha}{2}}}{E})^2 \times \hat{p} \times (1 - \hat{p})\\\\$

$= (\frac{1.3722}{0.04})^2 \times 0.5 \times 0.5\\\\= (34.305)^2 \times 0.5 \times 0.5\\\\= 1,176.83 \times 0.5 \times 0.5\\\\= 1,176.83 \times 0.25\\\\=294.2075$

5 0

3 years ago