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Helga [31]
2 years ago
14

An artist makes necklaces. She packs each necklace in a small jewelry box that is134inches by 214 inches by 34 inch. A departmen

t store ordered 270 necklaces. The artist plans to ship the necklaces to the department store using flat-rate shipping boxes from the post office. Which of the flat-rate boxes should she use to minimize her shipping cost?
Mathematics
1 answer:
9966 [12]2 years ago
8 0

Answer:

Small flat rate box

Step-by-step explanation:

Since she sells necklaces we can assume that the question is missing spaces and the actual dimensions are 1 3/4inch by 2 1/4 inch by 3/4 inch. This being the case she can use the Small flat rate box from the local post office for her necklaces. This is the smallest flat rate packaging available aside from envelopes, but since the artist needs a 3dimensional box the envelopes would not work. This small flat rate box has the following dimensions  8 11⁄16″ x 5 7⁄16″ x 1 3⁄4″ it also can hold up to 70lbs which far exceeds the weight of the necklaces.

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Gnom [1K]
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using
<span>a^3 + b^3 = (a+b)(a^2 − ab + b^2)
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w^3 + 125z^3
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hope it helps
6 0
3 years ago
2. A teacher asked his class of 20 students, "What is your age? Their responses are shown in the dot plot below.
Karolina [17]

Answer:

Yes, average does represent the centre of this data.

Average = 15.9

Step-by-step explanation:

As this question is not complete and not correctly written. Let's try to find an answer for this question.

Total number of Students = 20 = n

and we are given this data:

14, 15, 17, 18, 16.

which is just five entries and total number of students are 20. 15 entries are missing in this questions. So, let's add our own values into this question.

New set of entries for 20 students.

1. 14

2. 15

3. 17

4. 18

5. 16

6. 19

7. 17

8. 16

9. 17

10. 17

11. 15

12. 20

13. 14

14. 17

15. 18

16. 19

17. 13

18. 20

19. 17

20. 16

So, let's calculate the average of these numbers.

Average = Sum of all entries/ Total number of entries.

Average = 318/20

Average = 15.9

So, the average age of all the students of class is 15.9.

Let's calculate median:

For median , data must be arranged in a order from least to greatest number. Median is a number from that data which is at half way. In case of odd numbers, it is easy to identify median because there is only one number. and in case of even numbers, median is the average of the two middle numbers. So, here we have even numbers and median is the average of two middle numbers = 17 + 17/2 = 17

Median = 17

Yes, average does represent the centre of this data.

8 0
3 years ago
Read 2 more answers
Do these ratios form a proportion?
fenix001 [56]
No

There is not a common factor
7 0
3 years ago
Read 2 more answers
The binomial 8x+6 is equivalent to​
andreyandreev [35.5K]

Using the Distributive Property

The answer would be 3.

2 *4x = 8x and 2 * 3 = 6, to get 8x 6

5 0
3 years ago
Read 2 more answers
The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg
julia-pushkina [17]

Answer:

Approximately 101\; \rm ft (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

  • Let \rm O denote the observer.
  • Let \rm A denote the top of the tree.
  • Let \rm R denote the base of the tree.
  • Let \rm B denote the point where line \rm AR (a vertical line) and the horizontal line going through \rm O meets. \angle \rm B\hat{A}R = 90^\circ.

Angles:

  • Angle of elevation of the base of the tree as it appears to the observer: \angle \rm B\hat{O}R = 51^\circ.
  • Angle of elevation of the top of the tree as it appears to the observer: \angle \rm B\hat{O}A = 57^\circ.

Let the length of segment \rm BR (vertical distance between the base of the tree and the base of the hill) be x\; \rm ft.

The question is asking for the length of segment \rm AB. Notice that the length of this segment is \mathrm{AB} = (x + 20)\; \rm ft.

The length of segment \rm OB could be represented in two ways:

  • In right triangle \rm \triangle OBR as the side adjacent to \angle \rm B\hat{O}R = 51^\circ.
  • In right triangle \rm \triangle OBA as the side adjacent to \angle \rm B\hat{O}A = 57^\circ.

For example, in right triangle \rm \triangle OBR, the length of the side opposite to \angle \rm B\hat{O}R = 51^\circ is segment \rm BR. The length of that segment is x\; \rm ft.

\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}.

Rearrange to find an expression for the length of \rm OB (in \rm ft) in terms of x:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}.

Similarly, in right triangle \rm \triangle OBA:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}.

Equate the right-hand side of these two equations:

0.810\, x \approx 0.649\, (x + 20).

Solve for x:

x \approx 81\; \rm ft.

Hence, the height of the top of this tree relative to the base of the hill would be (x + 20)\; {\rm ft}\approx 101\; \rm ft.

6 0
3 years ago
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