3qt=2.83906L and 15lb=6.80389kg and 7L=7.39682qt hope this help
9514 1404 393
Answer:
382 square units
Step-by-step explanation:
The central four rectangles down the middle of the net are 9 units wide, and alternate between 8 and 7 units high. Then the area of those four rectangles is ...
9(8+7+8+7) = 270 . . . square units
The rectangles making up the two left and right "wings" of the net are 8 units high and 7 units wide, so have a total area of ...
2×(8)(7) = 112 . . . square units
Then the area of the figure computed from the net is ...
270 +112 = 382 . . . square units
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<em>Additional comment</em>
You can reject the first two answer choices immediately, because they are odd. Each face will have an area that is the product of integers, so will be an integer. There are two faces of each size, so <em>the total area of this figure must be an even number</em>.
You may recognize that the dimensions are 8, 8+1, 8-1. Then the area is roughly that of a cube with dimensions of 8: 6×8² = 384. If you use these values (8, 8+1, 8-1) in the area formula, you find the area is actually 384-2 = 382. That area formula is A = 2(LW +H(L+W)).
1/2=3/6
3/6+1/6=4/6 goodluck
By elimination:
y = 3x - 1
y = 2x + 2
Subtract the second equation from the first
0 = x - 1
y = 2x + 2
Subtract the first equation from the second
0 = x - 1
y = x + 3
Subtract the first equation from the second again
0 = x - 1
y = 4
Subtract x from both sides of the first equation
- x = - 1
y = 4
Divide the first equation by (-1)
x = 1
y = 4
<h3>
So, the solution is x = 1 and y = 4 {or: (1, 4)}</h3>
The probability of getting all heads is 1 / 2^6 = 1/64 as there is only 1 event where this happens in a possible 2^6 = 64 events. It is the same as the probability of getting all tails. The probability of getting at least 1 head is 1 - p(all tails) = 63/64.
