Please find the area!!!

A production process fills containers by weight. Weights of containers are approximately normally distributed. Historically, the

forsale [732]

Answer:

option (c) n = 201

Step-by-step explanation:

Data provided in the question:

Standard deviation, s = 5.5 ounce

Confidence level = 99%

Length of confidence interval = 2 ounces

Therefore,

margin of error, E = (Length of confidence interval ) ÷ 2

= 2 ÷ 2

= 1 ounce

Now,

E = $\frac{zs}{\sqrt n}$

here,

z = 2.58 for 99% confidence interval

n = sample size

thus,

1 = $\frac{2.58\times5.5}{\sqrt n}$

or

n = (2.58 × 5.5)²

or

n = 201.3561 ≈ 201

Hence,

option (c) n = 201

5 0

3 years ago

Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$