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AVprozaik [17]
3 years ago
12

A copper wire has a mass of 9.09grams and a volume of 1.01cubic centimeters. Calculate its density.

Mathematics
1 answer:
balu736 [363]3 years ago
8 0

Answer: The density of copper should be the same in a wire as it is in a chunk,

so the answer is 8.9 g/cm3.

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How do you write 20/10 as a mixed number
jolli1 [7]
It does not equal a mixed number, but a whole number so write 2 instead of a mixed number.
5 0
3 years ago
(2,5) (7,9) (3,9) (5,8) is it a function
NeX [460]

Answer:

(2,5) (7,9) (3,9) (5,8)

The relation is a function.

4 0
3 years ago
Using the GCF you found in Part B, rewrite 56 + 96 as two factors. One factor is the GCF and the other is the sum of two numbers
Vaselesa [24]
What was the GCF you found in part B?

8 0
3 years ago
at the city Museum, child admission is $5.10 an adult mission is $8.70. On Friday, 136 tickets were sold for total sales of $970
wariber [46]

Answer:

x = 76 children's tickets sold

Step-by-step explanation:

Let x = the number of children's tickets sold

Let y = the number of adult tickets sold

x + y = 131

5.5x + 9.8y = 957    Solve the first equation for y

y = 131 - x        Multiply the second equation by 10

55x +98y = 9570     Substitute the above equation into this one.

55x + 98(131-x) = 9570   Use the distributive property

55x + 12838 - 98x = 9570   Solve the equation

-43x = -3268

x= 76 children's tickets sold

7 0
3 years ago
The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined for each wafer in a s
GalinKa [24]

Answer:

1.   a, 99%       b71%

2. 64%, 44%

3. Number of particles    0, 1, 2, 3, 4,  5, 6, 7,  8, 9, 10, 11, 12, 13, 14

frequencies:                 1, 2, 3, 12, 11, 15, 18, 10,  12, 4, 5, 3, 1, 2, 1

Relative frequencies  0.01, 0.02,0.03,0.12,0.11,0.15, 0.18,0.1, 0.12,0.04,                        0.05,0.03,0.01,0.02,0.01

Step-by-step explanation:

Number of particles: 0, 1, 2, 3, 4, , 5, 6, 7

Frequency: 1, 2, 3, 12, 11, 15, 18, 10

Number of particles: 8, 9, 10, 11, 12, 13, 14

Frequency: 12, 4, 5, 3, 1, 2, 1

we solve by saying that  . one of th smaples wafer is with  any particles. we get the proportion by subtracting 1 from the total  divided by 100

100-1/100

.99

at least five particles , then we can subtract the number of those with particles less than five from 100 , divided by 100

100-(11+12+3+2+1)=71/100

0.71 or 71%

b. those particles within 5 and 10 inclusive

15+18+10+12+4+5=64/100

0.64

0r 64%

five and 10 exclusively will be

18+10+12+4=44/100

0.44

44%

c. Draw an histogram. first we have frequency table first

Number of particles    0, 1, 2, 3, 4,  5, 6, 7,  8, 9, 10, 11, 12, 13, 14

frequencies:                 1, 2, 3, 12, 11, 15, 18, 10,  12, 4, 5, 3, 1, 2, 1

Relative frequencies  0.01, 0.02,0.03,0.12,0.11,0.15, 0.18,0.1, 0.12,0.04,                        0.05,0.03,0.01,0.02,0.01

Relative frequencies is derived by dividing each frequencies over the total frequencies

wecan say that it is unimodal or say that it is slightly positively skewed

4 0
3 years ago
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