Which function has the same domain as y = 2 StartRoot x EndRoot
1 answer:
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Answer:
Step-by-step explanation:
we know that
when
-------> complementary angles
so
Because
-------> complementary angles
In this problem the product of
is equal to
therefore
Answer:
$996
Step-by-step explanation:
The rectangular plot has an area that is the product of its length and width. We are given the width as 12 feet, and the area as 240 ft², so we can find the length from ...
... A = L×W
... 240 ft² = L×(12 ft)
... 240 ft²/(12 ft) = L = 20 ft
Opposite sides of the rectangle are the same length, so the cost of fence for a side of a given length will be the sum of the costs of the opposites sides.
The 12 ft side has one segment that is $18 per foot, and one that is $15 per foot. For the 20 ft sides, both are $15 per foot. Then the total cost can be figured from ...
... (12 ft)·($18/ft + $15/ft) + (20 ft)·($15/ft +$15/ft) = 12·$33 +20·$30 = $996
Answer:
The probabilities of each outcome are the following:
for X = 1 is 0.6
for X = 2 is 0.35
for X = 3 is 0.049
and for X = 4 is 0.001
Step-by-step explanation:
Let's consider X as the random variable for the sum of outcomes "S" exceeds 3, this is:
Let's now analyze and consider the ways that X equals the different values:
X = 1: I throw the dice and the result is: 4, 5 and 6.
Then: P(X=1) = P(4) + P(5) + P(6) = 0.2+0.1+0.3 = 0.6
Thus P(X=1) = 0.6
X = 2: The result after following the dice two times can be:
1 and 3, 1 an 4.... and so on until 1 and 6
2 and 2, 2 and 3... and so on until 2 and 6
3 and 1... and so on until 3 a 6
Then P(X=2) = P(1)xP(3,4,5,6) + P(2)xP(2....6) + P(3)x(1.....6)
Theory of Probability: Sum of all possible outcomes P(1)+P(2).......P(6) = 1
Then P(1......6) = 1
Then P(X=2) = P(1)x[1-P(1)-P(2)]+P(2)x[1-P(1)]+P(3) = 0.1x(1-0.1-0.2) + 0.2x(1-0.1) + 0.1 = 0.1 x 0.7 + 0.2 x 0.9 + 0.1 = 0.35
Thus P(X=2) = 0.35
X = 3: The result can be
1 and 1 and 2, 1 and 1 and 3.... until 1 and 1 and 6
1 and 2 and 1, 1 and 2 and 2..... until 1 and 2 and 6
2 and 1 and 1, 2 and 1 and 2.... until 2 and 1 and 6
Then P(X=3) = P(1)xP(1)xP(2....6) + P(1)xP(2)xP(1.....6) + P(1)xP(2)xP(1.....6)
P(X=3) = 0.1 x 0.1 x (1-0.1) + 0.1 x 0.2 x 1 + 0.2 x 0.1 x 1 = 0.01 x 0.9 + 0.2 + 0.2 = 0.049
Thus P(X=3) = 0.049
Finally, for X to be 4, I only have the following possibilities
1 and 1 and 1 and 1.... until 1 and 1 and 1 and 6
Then P(X=4) = P(1)xP(1)xP(1)xP(1.....6) = 0.1x0.1x0.1x1 = 0.001
Thus P(X=3) = 0.001