Which is correct?

Please help! Need answer to number 4

alekssr [168]

Answer: H, 1/21

Step-by-step explanation:

7 letters in October

1 b / 7 letters

2 o / 6 letters

1/7 * 2/6 = 2/42 = 1 / 21

5 0

2 years ago

Karen deposited $25 in the bank on Monday, $50 on Wednesday and $15 on Friday. On Saturday, she took out $40. Karen's original b

natulia [17]

In order to solve this problem, use sign conventions. Since deposit indicates cash inflow, that would take the positive sign. On the other hand, withdrawals take negative sign. We add these integers to the original balance. he solution is as follows:

$100 + $25 + $50 + $15 + (⁻$40) = $150

3 0

2 years ago

Given the coordinates below, what is the midpoint of AB? A(-6, -10) B(2,5)

madreJ [45]

Given :

$\:$

A(-6, -10)
B( 2, 5)

$\:$

$\gray{ \frak{The \: given \: two \: \: points \: are(x_{1 },y_{1})=(−6 , -10)and(x_{ 2 },y_{2} )=( 2,5 )\:}}$

$\:$

Let's solve by using midpoint formula :

$\:$

$\bf \boxed{\color{red}\frak{Midpoint \: \: Formula : {( \: x ,\:y \: ) = }( \frak{ \frac{x_{1 } + y_{1}}{2} , \frac{x_{2 } + y_{2}}{2} )}}}$

$\:$

$\large \gray{ \frak{( \: x ,\:y \: ) = }( \frak{ \frac{ -6 + 2}{2} , \frac{ - 10 + 5}{2} )}}$

$\:$

$\: \large \gray{ \frak{( \: x ,\:y \: ) = }( \frak{ \frac{ -4}{2} , \frac{ - 5}{2} )}}$

$\:$

$\large \gray{ \frak{( \: x ,\:y \: ) = }( \frak{ \cancel\frac{ -4}{2} , \cancel\frac{ - 5}{2} )}}$

$\: \:$

$\underline{ \boxed{ \large \red{ \frak{option \: d }( \frak{ - 2, - 2.5)}}}}✓$

Hope Helps! :)

6 0

1 year ago

Chen writes down a two digit number. He finds out that if he swapped the numbers, it would be three more than the 1/3 of the ori

inna [77]

Answer:

Step-by-step explanation:

From the information given,

Let the two digits that Chen writes down to be p and q

where,

q will be in the unit place and p will be in the tens place.

So, the value of the digit = 10p + q

So, if he swapped the number, he will have 10q + p

However,

the new number determined is thrice more than the 1/3 pf the original number.

i.e

the 1/3rd of the original number = $\dfrac{1}{3}(10p+q)$

= $\dfrac{10p+q}{3}$

thrice more than that will be = $\dfrac{10p+q}{3} + 3$

∴

$\dfrac{10p+q}{3} + 3 = 10q+p$

multiply through by 3, we have :

10p+q + 9 = 30q + 3p

collecting the like terms, we have:

10p - 3p +q -30q +9 = 0

7p - 29q +9 = 0

7p - 29q = -9

Therefore, p,q lie between 0 and 9

Therefore, the possibility of the original number is p = 7 and q = 2

8 0

3 years ago

A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x)equals72 com

solmaris [256]

Answer:

Part (A)

1. Maximum revenue: $450,000

Part (B)

2. Maximum protit: $192,500

3. Production level: 2,300 television sets

4. Price: $185 per television set

Part (C)

5. Number of sets: 2,260 television sets.

6. Maximum profit: $183,800

7. Price: $187 per television set.

Explanation:

0. Write the monthly cost and price-demand equations correctly:

Cost:

$C(x)=72,000+70x$

Price-demand:

$p(x)=300-\dfrac{x}{20}$

Domain:

$0\leq x\leq 6000$

1. Part (A) Find the maximum revenue

Revenue = price × quantity

Revenue = R(x)

$R(x)=\bigg(300-\dfrac{x}{20}\bigg)\cdot x$

Simplify

$R(x)=300x-\dfrac{x^2}{20}$

A local maximum (or minimum) is reached when the first derivative, R'(x), equals 0.

$R'(x)=300-\dfrac{x}{10}$

Solve for R'(x)=0

$300-\dfrac{x}{10}=0$

$3000-x=0\\\\x=3000$

Is this a maximum or a minimum? Since the coefficient of the quadratic term of R(x) is negative, it is a parabola that opens downward, meaning that its vertex is a maximum.

Hence, the maximum revenue is obtained when the production level is 3,000 units.

And it is calculated by subsituting x = 3,000 in the equation for R(x):

R(3,000) = 300(3,000) - (3000)² / 20 = $450,000

Hence, the maximum revenue is $450,000

2. Part (B) Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set. 

i) Profit(x) = Revenue(x) - Cost(x)

Profit (x) = R(x) - C(x)

$Profit(x)=300x-\dfrac{x^2}{20}-\big(72,000+70x\big)$

$Profit(x)=230x-\dfrac{x^2}{20}-72,000\\\\\\Profit(x)=-\dfrac{x^2}{20}+230x-72,000$

ii) Find the first derivative and equal to 0 (it will be a maximum because the quadratic function is a parabola that opens downward)

Profit' (x) = -x/10 + 230

-x/10 + 230 = 0

-x + 2,300 = 0

x = 2,300

Thus, the production level that will realize the maximum profit is 2,300 units.

iii) Find the maximum profit.

You must substitute x = 2,300 into the equation for the profit:

Profit(2,300) = - (2,300)²/20 + 230(2,300) - 72,000 = 192,500

Hence, the maximum profit is $192,500

iv) Find the price the company should charge for each television set:

Use the price-demand equation:

p(x) = 300 - x/20

p(2,300) = 300 - 2,300 / 20

p(2,300) = 185

Therefore, the company should charge a price os $185 for every television set.

3. Part (C) If the government decides to tax the company $4 for each set it produces, how many sets should the company manufacture each month to maximize its profit? What is the maximum profit? What should the company charge for each set?

i) Now you must subtract the $4 tax for each television set, this is 4x from the profit equation.

The new profit equation will be:

Profit(x) = -x² / 20 + 230x - 4x - 72,000

Profit(x) = -x² / 20 + 226x - 72,000

ii) Find the first derivative and make it equal to 0:

Profit'(x) = -x/10 + 226 = 0

-x/10 + 226 = 0

-x + 2,260 = 0

x = 2,260

Then, the new maximum profit is reached when the production level is 2,260 units.

iii) Find the maximum profit by substituting x = 2,260 into the profit equation:

Profit (2,260) = -(2,260)² / 20 + 226(2,260) - 72,000

Profit (2,260) = 183,800

Hence, the maximum profit, if the government decides to tax the company $4 for each set it produces would be $183,800

iv) Find the price the company should charge for each set.

Substitute the number of units, 2,260, into the equation for the price:

p(2,260) = 300 - 2,260/20

p(2,260) = 187.

That is, the company should charge $187 per television set.

7 0

2 years ago