A charity sold 475 tickets to a dinner auction. If the charity raised 16,625 in ticket sales, what was the cost of one ticket?

The length of a kitchen wall is 24 2/3 feet long. A border will be placed along the wall of the kitchen. If the border strips ar

krek1111 [17]

Total length of the kitchen wall = 24 2/3 feet.

Let us convert 24 2/3 feet into improper fracion.

24 2/3 = (24*3+2)/3 = 74/3 feet.

Length of each strip = 1 3/4 feet.

Let us convert 1/3/4 feet in improper fracion.

1 3/4 feet = (1*4+3)/4 = 7/4 feet.

Total number of strips required = (Total length of the kitchen wall) ÷ (Length of each strip)

= 74/3 feet ÷ 7/4 feet.

Let us convert division sign into multiplication, we get

74/3 × 4/7.

Multiplying across, we get

296/21.

On dividing 296 by 21, we get 14.09 approximately.

14.09 represents more than 14.

So, we can round it to the next whole number, that is 15.

Therefore, 15 strips of border are needed.

5 0

3 years ago

3. In a single growing season at the Smith Family Orchard, the average yield per apple tree is 150 apples when the number of tre

g100num [7]

Answer:

A.10000

B.25 more trees must be planted

Step-by-step explanation:

⇒Given:

The intial average yield per acre $y_{i}$ = 150

The initial number of trees per acre $t_{i}$ = 100

For each additional tree over 100, the average yield per tree decreases by 1 i.e , if the number trees become 101 , the avg yield becomes 149.
Total yield = (number of trees per acre) $*$ (average yield per acre)

A.

⇒If the total trees per acre is doubled , which means :

total number of trees per acre $t_{f}$ = $2*t_{i}$ = 200

the yield will decrease by : $t_{f}$ - $y_{i}$

$y_{f}= 150-100= 50$

⇒total yield = $50*200=10000$

B.

⇒to maximize the yield ,

let's take the number of trees per acre to be 100+y ;

and thus the average yield per acre = 150 - y;

total yield = $(100+y)*(150-y)\\=15000+50y-y^{2} \\$

this is a quadratic equation. this can be rewritten as ,

⇒ $=15000+50y-y^{2}\\=15000+625 - (625 - 50y +y^{2})\\=15625 - (y-25)^{2}$

In this equation , the total yield becomes maximum when y=25;

⇒Thus the total number of trees per acre = 100+25 =125;

3 0

3 years ago

A soup can and label are shown. The label wraps around the circumference of the cylindrical can.

den301095 [7]

Answer: i would say 60 cm sorry if im wrong i think your missing a number

Step-by-step explanation:

4 0

3 years ago

How do I solve this and what kind of formula is it?

soldier1979 [14.2K]

Answer:

160

Step-by-step explanation:

The measure of an arc is always twice the degree of the corresponding angle inside the circle. For example, the measure of arc DC is just double the measure of angle DOC; angle DOC is 44 degrees, so arc DC must be 88 degrees.

Angle COB is 80 degrees, so arc CB is 160 degrees.

5 0

3 years ago

A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x)=74,000+80x

SOVA2 [1]

Answer:

a) $675000

b) $289000 profit,3300 set, $190 per set

c) 3225 set, $272687.5 profit, $192.5 per set

Step-by-step explanation:

a) Revenue R(x) = xp(x) = x(300 - x/30) = 300x - x²/30

The maximum revenue is at R'(x) =0

R'(x) = 300 - 2x/30 = 300 - x/15

But we need to compute R'(x) = 0:

300 - x/15 = 0

x/15 = 300

x = 4500

Also the second derivative of R(x) is given as:

R"(x) = -1/15 < 0 This means that the maximum revenue is at x = 4500. Hence:

R(4500) = 300 (4500) - (4500)²/30 = $675000

B) Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 80x) = -x²/30 + 300x - 80x - 74000

P(x) = -x²/30 + 220x - 74000

The maximum revenue is at P'(x) =0

P'(x) = - 2x/30 + 220= -x/15 + 220

But we need to compute P'(x) = 0:

-x/15 + 220 = 0

x/15 = 220

x = 3300

Also the second derivative of P(x) is given as:

P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3300. Hence:

P(3300) = -(3300)²/30 + 220(3300) - 74000 = $289000

The price for each set is:

p(3300) = 300 -3300/30 = $190 per set

c) The new cost is:

C(x) = 74000 + 80x + 5x = 74000 + 85x

Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 85x) = -x²/30 + 300x - 85x - 74000

P(x) = -x²/30 + 215x - 74000

The maximum revenue is at P'(x) =0

P'(x) = - 2x/30 + 215= -x/15 + 215

But we need to compute P'(x) = 0:

-x/15 + 215 = 0

x/15 = 215

x = 3225

Also the second derivative of P(x) is given as:

P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3225. Hence:

P(3225) = -(3225)²/30 + 215(3225) - 74000 = $272687.5

The money to be charge for each set is:

p(x) = 300 - 3225/30 = $192.5 per set

When taxed $5, the maximum profit is $272687.5

3 0

3 years ago

A charity sold 475 tickets to a dinner auction. If the charity raised 16,625 in ticket sales, what was the cost of one ticket?​

A charity sold 475 tickets to a dinner auction. If the charity raised 16,625 in ticket sales, what was the cost of one ticket?