Pls answer this with an explanation if you can pg3 pt1
2 answers:
You might be interested in
1) P= 64 cm, width = x, length = 14 + x
64 = x + x + 14 + x + 14 + x
64 = 4x + 28
4x = 64 - 28
4x = 36
x = 36/ 4
x = 9
AREA = 9 x (9+14)
Area = 9 x 23
Area = 207 cm
2) smaller no. = y, bigger no. = x
x - y = 3
2x+1/2y=36
Next, multiply 2 to first equation and find y:
2x-2y=6
5/2y=30
y = 12
Substitue y to x-y=3 to find x,
x = 15
3) x = pencil, y = pen
5x + 4y = $6.15
8x + 3y = 9.85$
8x + 7y = ?
5x + 4y = 6.15$ multiply both sides by -3
8x + 3y = 9.85$ multiply both sides by 4
-15x + 32x = (-3)*6.15 + 4*9.85
7x = 19.87 multiply both sides by 8/7
8x = (8/7) x 19.87
Add both equations:
32y - 15y = 8x6.15 - 5*9.58
17y = 1.3
y = 91/17) x 1.3
7y = (7/17)x 1.3
8x + 7y = (8/7) x 19.87 + (7/17) x 1.3
8x + 7y = $23.244
4) 2l + 2w = 0
2 x 2l + 2 x 1/2w = 40 + 16
l+w = 20
4l+w = 56
I’ve got l = 12 and w = 8
5) answer: 28.56 feet (approx)
step-by-step explanation:
since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.
we have to find out the measurement of segment ab.
therefore, tan 55° = \frac{ab}{bc}
but bc = 20 feet.
⇒ tan 55 ° = \frac{ab}{20}
⇒ ab = 20 × tan 55 °
⇒ ab = 20 × 1.42814800674
⇒ ab = 28.5629601348 ≈ 28.56 feet
6) The answer is teh attachment !
7) 3x-x+2=4
HOPE THE ANSWERS HELP!! PL MARK ME BRAINLIEST!
Let be the randomly selected lengths. Without loss of generality, suppose
where are independent random variables with the same uniform distribution on [0, 1].
By their mutual independence, we have
so that the joint density function is
where is the cube with vertices at (0, 0, 0) and (1, 1, 1).
Consider the plane
with . This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)
The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region , since
. So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.
Then we have
