Ask question

Ask question

All categories

2 years ago

9

A sporting goods store manager was selling a ski set for a certain price. The manager offered the markdowns shown, making the

one-day sale price of the ski set $325. Find the original selling price of the ski set. It was marked down 10% and 30%

1 answer:

Ilya [14]2 years ago

7 0

Answer:

The original selling price would be $ 515.87 ( approx )

Step-by-step explanation:

Let x be the original selling price ( in dollars ),

After marking down 10%,

New selling price = x - 10% of x = x - 0.1x = 0.9x

Again after marking down 30%,

Final selling price = 0.9x - 30% of 0.9x

= 0.9x - 0.3 × 0.9x

= 0.9x - 0.27x

= 0.63x

According to the question,

0.63x = 325

$\implies x =\frac{325}{0.63}\approx 515.87$

Hence, the original selling price would be $ 515.87.

You might be interested in

Find the geometric mean between 8 and 18. Write answer as a simplified radical. No Decimals!!!!

Crank

Answer: Therefore, the geometric mean of 8 and 18 is 12. Therefore, the geometric mean of 8 and 18 is 12.

Step-by-step explanation:

4 0

3 years ago

Need help ASAP due in 30 minutes!

Whitepunk [10]

Y=3x-4

Hope you get it turned in in time!!!

7 0

3 years ago

Read 2 more answers

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

$x \approx 1.548\,in$

The surface area of the box is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

4 0

2 years ago

Cheer100cheer100cheer100cheer100cheer100

dedylja [7]

Answer:

cool

Step-by-step explanation:

4 0

3 years ago

a rectangular garden is 7 feet long and 6 feet wide. a second rectangular garden has 3/2 the dimensions of the first garden .wha

Travka [436]

Answer:

125%

Step-by-step explanation:

Given: First Rectangular garden has dimension of 7 feet long and 6 feet wide.

Second Rectangular garden B has 3/2 the dimensions of the first garden.

Considerdering first garden as "A" and second garden as "B".

First lets find the Area for first rectangular garden.

Formula; Area of rectangle= $length\times width$

Area of Graden A= $7\times 6= 42\ feet$

∴ Area of Garden A is 42 feet.

Now, finding area of Garden B.

As given second rectangular garden has 3/2 the dimensions of the first garden.

∴ Length= $\frac{3}{2}\times 7= \frac{21}{2}\ feet$

Width= $\frac{3}{2} \times 6= 9\ feet$

∴ Area of Garden B= $\frac{21}{2} \times 9= \frac{189}{2}$

Area of Garden B= 94.5\ feet

Next, finding percent change from the first garden to the second garden.

Difference in area of garden from A to B= $94.5\ feet- 42\ feet= 52.5\ feet$

Percent change= $\frac{difference}{Area\ of\ first\ area}\times 100$

⇒ Percent change= $\frac{52.5}{42} \times 100= 125\%$

Hence, the percent in change from the first garden to the second garden is 125%.

4 0

3 years ago