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4 years ago

If they work together, Mark and Yvonne can sort 50 library books in 1.5 hours. Working alone, it would take

Mathematics

2 answers:

lozanna [386]4 years ago

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Answer:

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plz mark me as brainliest

Soloha48 [4]4 years ago

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(Awarding brainless) for the first correct answer

Step2247 [10]

Answer:B or A is the answer

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Please help me with the below question.

Snezhnost [94]

6a. By the convolution theorem,

$L\{t^3\star e^{5t}\} = L\{t^3\} \times L\{e^{5t}\} = \dfrac6{s^4} \times \dfrac1{s-5} = \boxed{\dfrac5{s^4(s-5)}}$

6b. Similarly,

$L\{e^{3t}\star \cos(t)\} = L\{e^{3t}\} \times L\{\cos(t)\} = \dfrac1{s-3} \times \dfrac s{1+s^2} = \boxed{\dfrac s{(s-3)(s^2+1)}}$

7. Take the Laplace transform of both sides, noting that the integral is the convolution of $e^t$ and $f(t)$ .

$\displaystyle f(t) = 3 - 4 \int_0^t e^\tau f(t - \tau) \, d\tau$

$\implies \displaystyle F(s) = \dfrac3s - 4 F(s) G(s)$

where $g(t) = e^t$ . Then $G(s) = \frac1{s-1}$ , and

$F(s) = \dfrac3s - \dfrac4{s-1} F(s) \implies F(s) = \dfrac{\frac3s}{\frac{s+3}{s-1}} = 3\dfrac{s-1}{s(s+3)}$

We have the partial fraction decomposition,

$\dfrac{s-1}{s(s+3)} = \dfrac13 \left(-\dfrac1s + \dfrac4{s+3}\right)$

Then we can easily compute the inverse transform to solve for f(t) :

$F(s) = -\dfrac1s + \dfrac4{s+3}$

$\implies \boxed{f(t) = -1 + 4e^{-3t}}$

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2 years ago

Suppose the parent population has an exponential distribution with a mean of 15 and standard deviation of 12. Use the Central Li

bazaltina [42]

Answer:

The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .