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Usimov [2.4K]
3 years ago
15

Please solve these coin problems:

Mathematics
1 answer:
ratelena [41]3 years ago
6 0
1)
Solving from system of equations:
x- amount of 1 cent stamps
y- amount of 33 cent stamps
\left \{ {{1*x+33*y=18,40} \atop {x+y=80}} \right.\\x=80-y\\0,01*(80-y)+0,33y=18,40\\0,8-0,01y+0,33y=18,40\\0,32y=18,40-0,8\\0,32y=17,60 /:0,32\\y=55\\x=80-55=25
2)
x- amount of 20 cent stamops
y-amount of 34 cent stamps
From system of equations:
\left \{ {{0,2x+0,34y=19,90} \atop {x+y=75}} \right.\\x=75-y\\0,2(75-y)+0,34y=19,90\\ 15-0,2y+0
,34y=19,90\\-0,2y+0,34y=19,90-15\\0,14y=4,90/:0,14\\ y=35\\x=75-35=40
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Find the circumference leave the answer in terms of pi
Sladkaya [172]
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So, C = 2 x 3.14 x 20

The answer is 40pi.


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4 years ago
I need help answering this
Lapatulllka [165]

Answer:

False

Step-by-step explanation:

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2 years ago
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Evaluate 10C4. Type a numerical answer in the space provided. Do not type spaces in your answer.
SOVA2 [1]
10C4 represent the Combinations of 10 objects taken 4 at a time. 

The formula for nCr is:

nCr= \frac{n!}{r!(n-r)!}

In the given case, n=10 and r=4. Using these values we get:

10C4= \frac{10!}{4!(10-4)!} \\  \\ 
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4 0
3 years ago
How do I find the Q1 and Q3?<br><br> 0,0,1,2,2,3,4,4,4,4,5,6,6,7,7
Angelina_Jolie [31]

Answer:

Q1 = 2

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Step-by-step explanation:

Mathematically, we have

Q1 = (n + 1)/4 th term

where n is the number of terms

By the count, we have n as 15

Q1 = (15 + 1)/4

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Looking at the arrangement, the 4th term is 2

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Q3 = 3(n + 1)/4 th term

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5 0
3 years ago
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