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Arturiano
2 months ago
6

A discount of 16% saved Jackie $38 on office supplies. What was the price of the supplies before the discount?

Mathematics
2 answers:
PIT_PIT [12.4K]2 months ago
8 0
You have two kinds of data $ and %!
Align them properly and perform cross multiplication.

16% corresponds to $38 as
100% corresponds to $x

Following this, x = (38*100)/16 = $237.5, which is the original price before the discount.
AnnZ [12.3K]2 months ago
3 0

Conclusion: The cost of the supplies prior to the discount is $237.5.

Step-by-step breakdown:  Given that Jackie saved $38 on office supplies due to a 16% discount.

We need to find the price of the supplies before the discount.

Let $x represent the pre-discount price of the supplies.

Thus, according to the information provided, we can say that

16\%\times x=38\\\\\\\Rightarrow \dfrac{16}{100}\times x=38\\\\\Rightarrow 16x=3800\\\\\Rightarrow x=\dfrac{3800}{16}\\\\\Rightarrow x=237.5.

Consequently, the price of the supplies before the discount is $237.5.

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Which of the following is true regarding the solutions to the logarithmic equation below? 2 log Subscript 6 Baseline (x) = 2. lo
zzz [12365]
Regarding the logarithmic equation, option c is correct: x=6 is a valid solution, while x=-6 is considered an extraneous solution. Explanation: The equation requires dividing both sides by 2, simplifying further until each step can be logically followed based on logarithmic definitions, from which we affirm that both sides equal when substituting back the values, confirming that x=6 is the true solution.
4 0
2 months ago
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Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th
Zina [12379]

Answer:

Step-by-step reasoning:

(a)

To have an accepted bid, it needs to exceed $10,000. Let bid x be a continuous random variable uniformly distributed between

$10,000 and $15,000

The range for accepted bidding is [ {\rm{\$ 10,000, \$ 15,000}], with b = $15000 and a = $10000.

The provided bidding range is [$10,000,$12,000]. The probability is determined as follows,

\begin{array}{c}\\P\left( {X{\rm{ < 12,000}}} \right){\rm{ = }}1 - P\left( {X > 12000} \right)\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{12000}^{15000}\\\end{array}

=1- \frac{[15000-12000]}{5000}\\\\=1-0.6\\\\=0.4

(b)  The accepted bidding range is [$10,000,$15,000], where b = $15,000 and a =$10,000. The given bidding range is [$10,000,$14,000].

\begin{array}{c}\\P\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{14000}^{15000}\\\end{array} P(X14000)

=1- \frac{[15000-14000]}{5000}\\\\=1-0.2\\\\=0.8

(c)

The optimal amount to bid for maximizing the probability of acquiring the property is calculated as,  

The accepted bidding range is [$10,000,$15,000],

where b = $15,000 and a = $10,000. The provided bidding range is [$10,000,$15,000].

\begin{array}{c}\\f\left( {X = {\rm{15,000}}} \right){\rm{ = }}\frac{{{\rm{15000}} - {\rm{10000}}}}{{{\rm{15000}} - {\rm{10000}}}}\\\\{\rm{ = }}\frac{{{\rm{5000}}}}{{{\rm{5000}}}}\\\\{\rm{ = 1}}\\\end{array}

(d)  If you know someone willing to pay you $16,000 for the property, would you still consider bidding less than the amount mentioned in part (c)? Why or why not?

5 0
2 months ago
A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150
lawyer [12517]

Answer:

Part 1) The equation is A(x)=150x-2x^2

Part 2) When x=40 m, the area of the schoolyard is A=2,800 m^2

Part 3) The valid domain consists of all real numbers exceeding zero and below 75 meters

Step-by-step explanation:

Part 1) Formulate an expression for A(x)

Let

x -----> the length of the rectangular school yard

y ---> the width of the rectangular school yard

It is known that

The perimeter for the fencing (taking the school wall as one side) is

P=2x+y

P=150\ m

thus

150=2x+y

y=150-2x -----> this is equation A

The area of the rectangular school yard is

A=xy ----> this is equation B

Substituting equation A into equation B yields

A=x(150-2x)

A=150x-2x^2

Change to function notation

A(x)=150x-2x^2

Part 2) What is the area when x=40?

With x equal to 40 m

substitute the value into the expression from Part 1 to determine A

A(40)=150(40)-2(40^2)

A(40)=2,800\ m^2    

Part 3) What would be a suitable domain for A(x) in this scenario?

We understand that

A signifies the area of the rectangular school yard

x characterizes the length of the rectangular school yard

It follows that

A(x)=150x-2x^2

This forms a vertical parabola opening downwards

The vertex indicates a maximum point

The x-coordinate of the vertex corresponds to the length that maximizes the area

The y-coordinate of the vertex denotes the maximum area

The vertex corresponds to (37.5, 2812.5)

Refer to the accompanying figure

Consequently,

The peak area achieved is 2,812.5 m^2

The x-intercepts are located at x=0 m and x=75 m

The domain for A is the range -----> (0, 75)

All real numbers greater than zero and less than 75 meters

5 0
3 months ago
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