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musickatia [10]
3 years ago
5

Jane downloads album A.

Mathematics
1 answer:
Masteriza [31]3 years ago
6 0

Answer:

a) total cost of downloading B is £11.90

b) total cost of downloading B is £12.04

see if that looks right lol

Step-by-step explanation:

a) £9.35 = 935p

935÷11=85p (cost per track)

85 x 14 = 1190p

1190p = £11.90

b) 85 x 6 = 510p

7 x 6 = 42p

510 - 42 = 468p

85 x 8 = 680p

7 x 8 = 56

680 + 56 = 736p

468 + 736 = 1204p

1204p = £12.04

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Here have some points
kozerog [31]

Answer:

Cheers bro ;)

5 0
3 years ago
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Carlos is using a number line to add another integer to –4. He begins by showing –4 on the number line, as shown below. Carlos a
Serhud [2]
<span>I had the same question on my quiz
The answer is that The number could be 3 or any integer less than 3.</span>
7 0
3 years ago
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The expression, involving exponents , to represent the shaded area, in square inches, diagram. Then use that expression to calcu
mart [117]

Answer:

The shaded area is 23\ in^{2}

Step-by-step explanation:

we know that

The shaded area is equal to the area of the large square minus the area of the two smaller squares

so

A=6^{2} -(3^{2} +2^{2})\\ \\ A=(2*3)^{2} -(3^{2} +2^{2})\\ \\A=(2^{2})(3^{2}) -(3^{2} +2^{2})

Calculate the shaded area

Remember that

3^{2}=9\\ 2^{2}=4

substitute

A=(4)(9) -(9 +4)\\ \\A=36-13\\ \\A=23\ in^{2}

5 0
3 years ago
For a process, the average range for all samples was 5 and the process average was 25. If the sample size was 10, calculate UCL
alexira [117]

Answer:

8.885

Step-by-step explanation:

Given that :

Sample size, n = 10

The average range, Rbar for all samples = 5

The upper control limit, UCL for the R-chart is :

UCL L= D4Rbar

From the control chart constant table, D4 = 1.777

Hence,

UCL = 1.777 * 5

UCL = 8.885

The UCL for the R-chart is 8.885

7 0
3 years ago
Which graph represents the solution set of the system of inequalities?
Novay_Z [31]
Answer:
(see image)
bottom right image

Explanation:
First try the origin (0,0) to rule out two of the graphs. 
3y ≥ x - 9 
3(0) ≥ (0) - 9 
3 ≥ - 9  yes 
3x + y > - 3 
3(0) + (0) > - 3
3 > - 3 yes 
so the origin should be in the shaded area of the graph, which rules out the top right and bottom left graphs. 
Now try a coordinate that is in the shaded area of one of the remaining graphs, but not in the other one. If it works, the graph is the one that has that point in the shaded region, and vice versa. 
Try point (4, 2)
3y ≥ x - 9 
3(2) ≥ (4) - 9
6 ≥  - 5 yes
3x + y > - 3 
3(4) + (2) > - 3
12 + 2 > - 3
14 > - 3 yes
So the graph is the bottom right one since (4, 2) is included in that shaded region.

4 0
3 years ago
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