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Sophie [7]
2 years ago
13

A high school coach needs to buy new athletic shorts for the 15 members of the basketball team. The coach must spend less than $

200 and needs to determine how much he can spend per pair of shorts. Write and solve an inequality to determine the maximum price for each pair of shorts. What does the solution represent?
Mathematics
2 answers:
Monica [59]2 years ago
7 0

Answer: The coach may spend up to $13.33 per pair of shorts.

Step-by-step explanation:

Hi, to answer this question we have to write an inequality with the information given:

  • <em>Number of shorts: 15 (for 15 members) </em>
  • <em>Budget: $200 </em>

So, we have to multiply the number of shorts by the price of each one, we will represent the price with the variable "x".(15x)

That cost must be equal or less to 200.

Mathematically speaking

15 x ≤ 200

Solving for x

x ≤200/15

x ≤ 13.33

This solution represents that the coach may spend up to $13.33 per pair of shorts.

Feel free to ask for more if needed or if you did not understand something.

Studentka2010 [4]2 years ago
4 0

Answer:

15x < 200; x < 13.33; the maximum price for a pair of shorts

Step-by-step explanation:

1. Set up the inequality

Let x = price of a pair of shorts. Then

 15x = price of shorts for the team

You have one condition:

15x < 200

2. Solve the inequality

\begin{array}{rcl}15x & < & 200\\\\x & < & \dfrac{200}{15}\\\\x & < & \mathbf{13.333}\\\end{array}

3. Meaning of solution

The solution represents the maximum price the coach can pay for a pair of shorts.

If the coach pays $13.33 per pair, the total cost for the team will be $199.95, and the condition is satisfied.

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In a local ice sculpture contest, one group sculpted a block into a rectangular based pyramid. The dimensions of the base were 3
m_a_m_a [10]

Answer:

1. The amount of ice needed = 18 m²

2. The amount of fabric needed to manufacture the umbrella is 0.76 m²

3. The height of the cone, is 3.75 cm

4. The dimensions of the deck are;

Width = 28/3 m, breadth = 28/3 m

The area be 87.11 m²

5.   The dimensions of the optimal design for setting the storage area at the corner, we have;

Width = 10m

Breadth = 10 m

The dimensions of the optimal design for setting the storage area at the back of their building are;

Width = 7·√2 m

Breadth = 7·√2 m

Step-by-step explanation:

1. The amount of ice needed is given by the volume, V, of the pyramid given by V = 1/3 × Base area × Height

The base area = Base width × Base breadth = 3 × 5 = 15 m²

The pyramid height = 3.6 m

The volume of the pyramid = 1/3*15*3.6 = 18 m²

The amount of ice needed = 18 m²

2. The surface area of the umbrella = The surface area of a cone (without the base)

The surface area of a cone (without the base) = π×r×l

Where:

r = The radius of the cone = 0.4 m

l = The slant height = √(h² + r²)

h = The height of the cone = 0.45 m

l = √(0.45² + 0.4²) = 0.6021 m

The surface area = π×0.4×0.6021 = 0.76 m²

The surface area of a cone (without the base) = 0.76 m²

The surface area of the umbrella = 0.76 m²

The amount of fabric needed to manufacture the umbrella = The surface area of the umbrella = 0.76 m²

3. The volume, V, of the cone = 1/3×Base area, A, ×Height, h

The volume of the cone V = 150 cm³

The base area of the cone A = 120 cm²

Therefore we have;

V = 1/3×A×h

The height of the cone, h = 3×V/A = 3*150/120 = 3.75 cm

4. Given that the deck will have railings on three sides, we have;

Maximum dimension = The dimension of a square as it is the product of two  equal maximum obtainable numbers

Therefore, since the deck will have only three sides, we have that the length of each side are equal and the fourth side can accommodate any dimension of the other sides giving the maximum dimension of each side as 28/3

The dimensions of the deck are width = 28/3 m, breadth = 28/3 m

The area will then be 28/3×28/3 = 784/9 = 87\frac{1}{9} =87.11 m²

5. The optimal design for setting the storage area at the corner of their property with four sides is having the dimensions to be that of of a square with equal sides of 10 m each as follows;

Width = 10m

Breadth = 10 m

The optimal design to have the storage area at the back of their building having a fence on only three sides, is given as follows;

Storage area specified = 98 m²

For optimal use of fencing, we have optimal side size of fencing = s = Side length of a square

s² = 98 m²

Therefore, s = √98 = 7·√2 m

Which gives the width = 7·√2 m and the breadth = 7·√2 m.

8 0
1 year ago
A book costs d dollars.
bazaltina [42]

Answer:

The total amount paid by Jill  = $ (3 d)

Step-by-step explanation:

Here, the cost of 1 book = $ d

Now, the number of books purchased by Jill = 3

So, the total cost of 3 books  = 3 x ( Cost of 1 book)

                                                 = 3 x ($d)  = 3 d

So, if Jill purchased 3 books, the total amount paid by her = $ (3 d)

3 0
1 year ago
A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l
Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = <u><em>the length of the calls, in minutes.</em></u>

So, X ~ Normal(\mu=4.5,\sigma^{2} =0.70^{2})

The z-score probability distribution for the normal distribution is given by;

                           Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 4.5 minutes

           \sigma = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X \leq 4.50 min)

    P(X < 5.30 min) = P( \frac{X-\mu}{\sigma} < \frac{5.30-4.5}{0.7} ) = P(Z < 1.14) = 0.8729

    P(X \leq 4.50 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.5-4.5}{0.7} ) = P(Z \leq 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

    P(X > 5.30 min) = P( \frac{X-\mu}{\sigma} > \frac{5.30-4.5}{0.7} ) = P(Z > 1.14) = 1 - P(Z \leq 1.14)

                                                              = 1 - 0.8729 = <u>0.1271</u>

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 5.30 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 5.30 min) = P( \frac{X-\mu}{\sigma} \leq \frac{5.30-4.5}{0.7} ) = P(Z \leq 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X \leq 4.00 min)

    P(X < 6.00 min) = P( \frac{X-\mu}{\sigma} < \frac{6-4.5}{0.7} ) = P(Z < 2.14) = 0.9838

    P(X \leq 4.00 min) = P( \frac{X-\mu}{\sigma} \leq \frac{4.0-4.5}{0.7} ) = P(Z \leq -0.71) = 1 - P(Z < 0.71)

                                                              = 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

            P(X > x) = 0.05            {where x is the required time}

            P( \frac{X-\mu}{\sigma} > \frac{x-4.5}{0.7} ) = 0.05

            P(Z > \frac{x-4.5}{0.7} ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

                      \frac{x-4.5}{0.7}=1.645

                      {x-4.5}{}=1.645 \times 0.7

                       x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0
2 years ago
What is the completely factored form 8x2-50y2?
lorasvet [3.4K]
Remember
a²-b²=(a-b)(a+b)

8x²-50y²
2(4x²-25y²)
(2)([2x]²-[5y]²)
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7 0
2 years ago
Events M and N are independent events. In this scenario, if P(M) = 0.46 and P(M and N) = 0.138, then P(N) = .
Oksi-84 [34.3K]

Answer:

P(N)=0.3

Step-by-step explanation:

Given: P(M)= 0.46, P(M and N)=0.138

Using P(M) ×P(N)= P(M and N)

⇒0.46×P(N)= 0.138

⇒P(N)= \frac{0.138}{0.46}

⇒P(N)=0.3

7 0
2 years ago
Read 2 more answers
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