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Mice21 [21]
3 years ago
5

Unit cubes in solid figures

Mathematics
1 answer:
andrey2020 [161]3 years ago
6 0
So what you need to do is find the less he can use.
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Jackson bought snacks for his team's practice. He bought a bag of oranges for $1.61 and a 6-pack of juice bottles. The total cos
ki77a [65]

Answer:

$1.93

Step-by-step explanation:

From the total (13.19), <u>subtract</u> the cost of the oranges (1.61). That would equal 11.58, which is the cost of the 6-pack of juice bottles. Then, <u>divide </u>11.58 by 6 to determine the cost of each bottle.

Each bottle = $1.93

Hope this helps!!

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3 years ago
A frame is marked down by 22 percent from an original price of $14.50. What is the new price?
melamori03 [73]

Answer:

11.31

Step-by-step explanation:

5 0
3 years ago
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Beth has 3,000 feet of fencing available to enclose a rectangular field. One side of the field lies along a river, so only three
Yakvenalex [24]
A(x) = (3000-x)/2 * x
a(x)=(1500-½x)*x
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3 years ago
PLS HELP PLS I BEG U
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Answer:
Equation: y=4x







7 0
3 years ago
Let a "binary code" be the set of all binary words, each consisting of 7 bits (i.e., 0 or 1 digits). For example, 0110110 is a c
Dmitry_Shevchenko [17]

Answer:

a) 128 codewords

b) 35 codewords

c) 29 codewords

Step-by-step explanation:

a) Each 7 bits consist of 0 or 1 digits. Therefore the first bit is two choices (0 or 1), the second bit is also two choices (0 or 1), continues this way till the last bit.

So total number of different code words in 7 bits is 2×2×2×2×2×2×2 = 2⁷ = 128

There are 128 different codewords.

b) A code word contains exactly four 1's this means that  it has four 1's and three 0's . Therefore, in 7 bits, we have four of the same kind and three of the same kind. Hence, total number of code words containing exactly four 1's =7!/(4!*3!) = 35 codewords

c) number of code words containing at most two 1's  = codewords containing zero 1's + words containing one 1's + words containing two 1's

Now codewords containing zero 1's = 0000000 so 1 word

Codewords containing one 1's = 1000000,0100000,0010000,0001000,0000100,0000010,0000001. That's seven words

Codewords containing two 1's means word containing two 1's and five 0's. So out of seven, two are of one kind and five are of another kind

Therefore, the total number of such words=7!/(2!*5!)=21

Hence, codewords having at most two 1's = 21+7+1 =29 codewords

8 0
3 years ago
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