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3 years ago

Help me solve for x. 16^x = x^2

1 answer:

8 0

Hello! Let’s begin, But before that, I hope you can mark me as the brainliest, just a hope.

x * 2^4x = x^2

2^4x = x

I’m afraid you can’t simplify more, but it could be

2^2x/2 = root x

you could solve for it yourself, because the other guy is not likely to answer.

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Sales tax is 6%. Bought car for$820 what was cost of car

pogonyaev

820 + .06(820)
820 + 49.2
869.2
The cost of the car is $869.20.

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4 years ago

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A man travels 320 miles in 8 hours. If he continues at the same rate, how many miles will he

evablogger [386]

Answer: 80 miles

Step-by-step explanation:

320miles/8hours= 40mph

40mph=xmiles/2hrs

40*2= 80 miles

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3 years ago

The sum of three consecutive odd integers is − 87 . find the numbers.

DIA [1.3K]

Three consecutive odd integers are n, n+2 and n+4

n + n+2 + n+4 = -87
3n + 6 = -87
3n = -87 - 6
3n = -93
n = -93/3
n = -31

n+2 = -31 + 2 = -29
n+4 = -31 + 4 = -27

The numbers are -31, -29 and -27

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4 years ago

the cone below has a diameter of 12 units and a slant height of 10 units. Find the volume of the cone

Oduvanchick [21]

Answer: 376.8 units^3

Step-by-step explanation:

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4 years ago

(a) A lamp has two bulbs of a type with an average lifetime of 1600 hours. Assuming that we can model the probability of failure

lara [203]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability is $P_T= 0.4560$

The probability is $P_F= 0.0013$

Step-by-step explanation:

From the question we are told that

The mean for the exponential density function of bulbs failure is $\mu = 1600 \ hours$

Generally the cumulative distribution for exponential distribution is mathematically represented as

$1 - e^{- \lambda x}$

The objective is to obtain the p=probability of the bulbs failure within 1800 hours

So for the first bulb the probability will be

$P_1(x < 1800)$

And for the second bulb the probability will be

$P_2 (x< 1800)$

So from our probability that we are to determine the area to the left of 1800 on the distribution curve

Now the rate parameter $\lambda$ is mathematically represented as

$\lambda$ $= \frac{1}{\mu}$

$\lambda = \frac{1}{1600}$

The probability of the first bulb failing with 1800 hours is mathematically evaluated as

$P_1(x < 1800) = 1 - e^{\frac{1}{1600} * 1800 }$

$= 0.6753$

Now the probability of both bulbs failing would be

$P_T=P_1(x < 1800) * P_2(x < 1800)$

$= 0.6375 * 06375$

$P_T= 0.4560$

Let assume that one bulb failed at time $T_a$ and the second bulb failed at time $T_b$ then

$T_a + T_b = 1800\ hours$

The mathematical expression to obtain the probability that the first bulb failed within between zero and $T_a$ and the second bulb failed between $T_a \ and \ 1800$ is represented as

$P_F=\int_{0}^{1800}\int_{0}^{1800-x} \f{\lambda }^{2}e^{-\lambda x}* e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\lambda }e^{-\lambda x}\int_{0}^{1800-x} {\lambda } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}\int_{0}^{1800-x} \frac{1}{1600 } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}[e^{- \lambda y}]\left {1800-x} \atop {0}} \right. dx$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\frac{x}{1600} }[e^{- \frac{1800 -x}{1600} }-1] dx$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{x}{1600} }] \left {1800} \atop {0}} \right.$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{1800}{1600} }] -[[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{-0}]$

$=[\frac{1}{1600} e^{-\frac{1800}{1600} } - \frac{1}{1600} e^{-0} ]$

$=0.001925 -0.000625$

$P_F= 0.0013$

4 0

3 years ago

Read 2 more answers

Help me solve for x. 16^x = x^2​

Help me solve for x. 16^x = x^2