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3 years ago

The average price of a laptop is $965. Assume laptop prices are approximately normally distributed with a standard

Mathematics

1 answer:

miss Akunina [59]3 years ago

6 0

Answer:

$836.8

Step-by-step explanation:

Average price = mean = $965

Standard deviation, = $100

Given that distribution is approximately normal ;

The least expensive 10% of the laptops :

We Obtain the Zscore that corresponds to P(Z ≤ 0.1) ; this means the least 10% of the laptops ;

From, a normal probability distribution table ;

P(Z ≤ 0.1) = - 1.282

We substitute this into the Zscore formula :

Zscore = (x - mean ) / standard deviation

x = price

-1.282 = (x - 965) / 100

-128.2 = (x - 965)

x = - 128.2 + 965

x = $836.8

Hence, price is $836.8

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Answer:

The maximum amount of red kryptonite present is 33.27 g after 4.93 hours.

Step-by-step explanation:

dy/dt = y(1/t - k)

separating the variables, we have

dy/y = (1/t - k)dt

dy/y = dt/t - kdt

integrating both sides, we have

∫dy/y = ∫dt/t - ∫kdt

㏑y = ㏑t - kt + C

㏑y - ㏑t = -kt + C

㏑(y/t) = -kt + C

taking exponents of both sides, we have

$\frac{y}{t} = e^{-kt + C} \\\frac{y}{t} = e^{-kt}e^{C} \\\frac{y}{t} = Ae^{-kt} (A = e^{C})\\y = Ate^{-kt}$

when t = 1 hour, y = 15 grams. So,

$y = Ate^{-kt}\\15 = A(1)e^{-kX1}\\15 = Ae^{-k}$ (1)

when t = 3 hours, y = 30 grams. So,

$y = Ate^{-kt}\\30 = A(3)e^{-kX3}\\30 = 3Ae^{-3k}$ (2)

dividing (2) by (1), we have

$\frac{30}{15} = \frac{3Ae^{-3k}}{Ae^{-k}} \\2 = 3e^{-2k}\\\frac{2}{3} = e^{-2k}$

taking natural logarithm of both sides, we have

-2k = ㏑(2/3)

-2k = -0.4055

k = -0.4055/-2

k = 0.203

From (1)

$A = 15e^{k} \\A = 15e^{0.203} \\A = 15 X 1.225\\A = 18.36$

Substituting A and k into y, we have

$y = 18.36te^{-0.203t}$

The maximum value of y is obtained when dy/dt = 0

dy/dt = y(1/t - k) = 0

y(1/t - k) = 0

Since y ≠ 0, (1/t - k) = 0.

So, 1/t = k

t = 1/k

So, the maximum value of y is obtained when t = 1/k = 1/0.203 = 4.93 hours

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<u>So the maximum amount of red kryptonite present is 33.27 g after 4.93 hours.</u>

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