Answer:
A non-equilateral rhombus.
Step-by-step explanation:
We can solve this graphically.
We start with square:
ABCD
with:
A = (11, - 7)
B = (9, - 4)
C = (11, - 1)
D = (13, - 4)
Only with the vertices, we can see that ABCD is equilateral, as the length of each side is:
AB = √( (11 - 9)^2 + (-7 -(-4))^2) = √( (2)^2 + (3)^2) = √(4 + 9) = √13
BC = √( (11 - 9)^2 + (-1 -(-4))^2) = √13
CD = √( (11 - 13)^2 + (-1 -(-4))^2) = √13
DA = √( (11 - 13)^2 + (-7 -(-4))^2) = √13
And we change C by C' = (11, 1)
In the image you can see the 5 points and the figure that they make:
The figure ABCD is a rhombus, and ABC'D is also a rhombus, the only difference between the figures is that ABCD is equilateral while ABC'D is not equilateral.
Answer: Zack won. If you compare the decimal equivalents of their scores, -3.666 < -3.625
Step-by-step explanation:
Given: Vanessa and Zack are playing a game where the player with the lower score wins.
Score of Vanessa=
Score of Zack =
We can see
⇒ Zack has the lower score .
Hence, Zack won. If you compare the decimal equivalents of their scores, -3.666 < -3.625.
Answer:
0.96 = 96% probability that at least one of them detect an enemy attack.
Step-by-step explanation:
For each radar, there are only two possible outcomes. Either it detects the attack, or it does not. The missiles are operated independently, which means that we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
Assume that a particular detection system has a 0.80 probability of detecting a missile attack.
This means that
If two military radars are installed in two different areas and they operate independently, the probability that at least one of them detect an enemy attack is
This is when
. So
In which
0.96 = 96% probability that at least one of them detect an enemy attack.