Michael's total sales last month

The management of Ditton Industries has determined that the daily marginal revenue function associated with selling x units of t

barxatty [35]

Answer:

(a) The daily total revenue realized from the sale of 210 units of the toaster oven is $6,195.

(b) The additional revenue realized when the production (and sales) level is increased from 210 to 310 units is $1,400.

Step-by-step explanation:

The marginal revenue gives the actual revenue realized from the sale of an additional unit of the commodity given that sales are already at a certain level. The derivative R' of the function R measures the rate of change of the revenue function.

We know that the daily marginal revenue function is given by

$R'(x) = -0.1x+40$

(a) To find the the daily total revenue you must:

Integrate the daily marginal revenue function,

$\int R'(x)dx = \int(-0.1x+40)dx\\\\R(x)=-\int \:0.1xdx+\int \:40dx\\\\R(x)=-0.05x^2+40x+C$ ,

where C is a constant.

Find the value of C, using the fact that if you sell 0 units your daily revenue is $0.

$0=-0.05(0)^2+40(0)+C\\C=0$

$R(x)=-0.05x^2+40x$

The daily total revenue realized from the sale of 210 units of the toaster oven is

x = 210 units

$R(210)=-0.05(210)^2+40(210)\\R(210)=-210^2\cdot \:0.05+8400\\R(210)=-2205+8400\\R(210)=6195$

(b) To find the additional revenue realized when the production (and sales) level is increased from 210 to 310 units you must:

Find the daily total revenue realized from the sale of 310 units

$R(310)=-0.05(310)^2+40(310)\\R(310)=-310^2\cdot \:0.05+12400\\R(310)=-4805+12400\\R(310)=7595$

The additional revenue realized when the production (and sales) level is increased from 210 to 310 units is

$R(310)-R(210)=7595-6195=1400$

3 0

4 years ago

The heights of 40 randomly chosen men are measured and found to follow a normal distribution. An average height of 175 cm is obt

AVprozaik [17]

Answer:

95% two-sided confidence interval for the true mean heights of men is [168.8 cm , 181.2 cm].

Step-by-step explanation:

We are given that the heights of 40 randomly chosen men are measured and found to follow a normal distribution.

An average height of 175 cm is obtained. The standard deviation of men's heights is 20 cm.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average height = 175 cm

$\sigma$ = population standard deviation = 20 cm

n = sample of men = 40

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ]

= [ $175-1.96 \times }{\frac{20}{\sqrt{40} } }$ , $175+1.96 \times }{\frac{20}{\sqrt{40} } }$ ]

= [168.8 cm , 181.2 cm]

Therefore, 95% confidence interval for the true mean height of men is [168.8 cm , 181.2 cm].

The interpretation of the above interval is that we are 95% confident that the true mean height of men will be between 168.8 cm and 181.2 cm.

3 0

3 years ago

Diga um numero divisivel por 539 e um de 3465

Gwar [14]

There are a few ways to find this. You can use prime factorization to find the LCM or the GCF

The GCF = 77 so 77 will divide into 539 and 3465

Answer = 77
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Hay algunas formas de encontrar esto. Puede usar factorización prima para encontrar el LCM o el GCF

El GCF = 77 así que 77 se dividirá en 539 y 3465

responder = 77

8 0

3 years ago

Differentiate from first principles y=3x

wel

$\frac{d}{dx} 3x = 3$