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Gelneren [198K]
3 years ago
6

a psychologist contends that the number of facts of a certain type that are remembered after t hours is given by the following f

unction. ​f(t)equals startfraction 85 t over 99 t minus 85 endfraction find the rate of change at which the number of facts remembered is changing after 1 hour and after 10 hours.
Mathematics
1 answer:
laiz [17]3 years ago
4 0

Answer:

At t=1, Rate of Change=-36.86

At t=10 hours, Rate of Change =-0.0088

Step-by-step explanation:

The function which describes the number of facts of a certain type which are remembered after t hours is given as:

f(t)=\frac{85t}{99t-85}

To determine the Rate of Change at the given time, we first look for the derivative of f(t).

Applying quotient rule:

f^{'}(t)=\frac{-7225}{{\left( 85 - 99\,t\right) }^{2}}

At t=1

f^{'}(1)=\frac{-7225}{(85-99)^{2}}

=-36.86

At t=10 hours

f^{'}(10)=\frac{-7225}{(85-99(10))^{2}}

=-0.0088

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The probability will be 1/6
7 0
3 years ago
Laine reads 25 pages in 30 minutes.
zlopas [31]

Answer:

4 hours

Step-by-step explanation:

25 * 8 = 200

so, 30 * 8 = 240(minutes)

240 / 60 = 4 hours

Hope it helps!

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7 0
3 years ago
A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and
mr Goodwill [35]

Answer:

Bias for the estimator = -0.56

Mean Square Error for the estimator = 6.6311

Step-by-step explanation:

Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.

To find - Determine the bias and the mean squared error for this estimator of the mean.

Proof -

Let us denote

X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)

Now,

An estimate of mean, μ is suggested as

\mu = \frac{3X_{1} + 4X_{2}  }{8}

Now

Bias for the estimator = E(μ bar) - μ

                                    = E( \frac{3X_{1} + 4X_{2}  }{8}) - 4.5

                                    = \frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5

                                    = \frac{3(4.5) + 4(4.5)}{8} - 4.5

                                    = \frac{13.5 + 18}{8} - 4.5

                                    = \frac{31.5}{8} - 4.5

                                    = 3.9375 - 4.5

                                    = - 0.5625 ≈ -0.56

∴ we get

Bias for the estimator = -0.56

Now,

Mean Square Error for the estimator = E[(μ bar - μ)²]

                                                             = Var(μ bar) + [Bias(μ bar, μ)]²

                                                             = Var( \frac{3X_{1} + 4X_{2}  }{8}) + 0.3136

                                                             = \frac{1}{64} Var( {3X_{1} + 4X_{2}  }) + 0.3136

                                                             = \frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})]  }) + 0.3136

                                                             = \frac{1}{64} [{3(57.76) + 4(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [7(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [404.32]  } + 0.3136

                                                             = 6.3175 + 0.3136

                                                              = 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311

6 0
3 years ago
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Answer: Experimental probability

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There are two kinds of probability: Theoretical probability and Experimental probability.

To calculate theoretical probability we divide favorable outcomes by total outcomes.

To calculate experimental probability we divide number of times an event occurs by the total number of trials or times the activity is performed.

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3 years ago
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PolarNik [594]

The solution is -4/9x + 1

7 0
3 years ago
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