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SVETLANKA909090 [29]
3 years ago
9

Marianne has been collecting donations for her biscuit stall at the school summer fayre. There are some luxury gift tins of bisc

uits to be sold at £5 each, normal packets at £1 each, and mini-packs of 2 biscuits at 10p each. She tells Amy that she has recieved exactly 100 donations in total, with a collective value of £100, amd that her stock of £1 packets is very low compared with the other items. Amy wants to work out how many of each item Marianne has. Show how she can do it.
Mathematics
2 answers:
sergeinik [125]3 years ago
7 0
The problem conditions give rise to 2 equations in 3 unknowns. Let L, M, N represent the number of Luxury, Mini, and Normal packets sold.
.. L +M +N = 100 . . . . . . the number of packets sold
.. 5L +0.1M +N = 100 . . the value of donations
These result in the relationships
.. L = (9/40)M
.. N = 100 -(49/40)M

There are three integer solutions in which the numbers are non-negative.
.. (L, M, N) = (0, 0, 100) or (9, 40, 51) or (18, 80, 2)

If Marianne sold 100 normal packets, her stock would be "very low" compared to the others.

If Marianne sold 51 normal packets, her stock may be "very low" with respect to the others, depending on how many of each she started with. This might be the solution if we require non-zero numbers of all packets were sold.
Marta_Voda [28]3 years ago
6 0
I don't know what 10p is so before I answer I'd have to know what that is
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nikitadnepr [17]
X = 77.789
y = 23...............................
4 0
3 years ago
Explain 15 is not a prime factor
MissTica

<em>1</em><em>5</em><em> </em><em>is</em><em> </em><em>not</em><em> </em><em>a</em><em> </em><em>prime</em><em> </em><em>factor</em><em> </em><em>as</em><em> </em><em>it</em><em> </em><em>is</em><em> </em><em>divided</em><em> </em><em>by</em><em> </em><em>1</em><em>,</em><em>3</em><em>,</em><em>5</em><em> </em><em>and</em><em> </em><em>1</em><em>5</em><em>.</em>

<em>Additional</em><em> </em><em>information</em><em>:</em>

<em>Prime</em><em> </em><em>numbers</em><em> </em><em>are</em><em> </em><em>those</em><em> </em><em>numbers</em><em> </em><em>which</em><em> </em><em>can</em><em> </em><em>only</em><em> </em><em>be</em><em> </em><em>divided</em><em> </em><em>by</em><em> </em><em>itself</em><em> </em><em>.</em><em>for</em><em> </em><em>instance</em><em>:</em><em>1</em><em>,</em><em>2</em><em>,</em><em>3</em><em>,</em><em>5</em><em>,</em><em>7</em><em> </em><em>,</em><em>1</em><em>1</em><em>,</em><em>1</em><em>3</em><em> </em><em>etc</em><em>.</em>

<em>Composite</em><em> </em><em>number</em><em> </em><em>s</em><em> </em><em>are</em><em> </em><em>those</em><em> </em><em>numbers</em><em> </em><em>which</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>also</em><em> </em><em>divided</em><em> </em><em>by</em><em> </em><em>other</em><em> </em><em>numbers</em><em>,</em><em>For</em><em> </em><em>instance</em><em>:</em><em> </em><em>4</em><em>,</em><em>6</em><em>,</em><em>8</em><em>,</em><em>1</em><em>0</em><em> </em><em>etc</em>

<em>Hope</em><em> </em><em>it </em><em>helps</em><em>.</em><em>.</em><em>.</em><em>.</em>

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=x%2B0.08%2B3.21%5Cleq%2060" id="TexFormula1" title="x+0.08+3.21\leq 60" alt="x+0.08+3.21\leq 6
Rashid [163]
There is no solution to this, due to X can be equal anything. Hope this helps!
8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%7B9%7D%5E%7B2%7D%20-%202%281%20%2B%203%29" id="TexFormula1" title=" {9}^{2} - 2(1 + 3)" al
Dmitrij [34]
The answer is 73 because you solve 9^ 2 which is 81 then multiply 2 times 4 which is 8 then subtract 81 and 8 which is 73
8 0
3 years ago
Read 2 more answers
Need answer quick. No need for big explanantion
Leona [35]

Answer:

Step-by-step explanation:

4.30/40=$0.1075

Hope this helps :)

5 0
3 years ago
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