Which point is a solution
2 answers:
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See the picture attached to better understand the problem
we know that
in the right triangle ABC
cos A=AC/AB
cos A=1/3
so
1/3=AC/AB----->AB=3*AC-----> square----> AB²=9*AC²----> equation 1
applying the Pythagoras Theorem
BC²+AC²=AB²-----> 2²+AC²=AB²---> 4+AC²=AB²----> equation 2
substitute equation 1 in equation 2
4+AC²=9*AC²----> 8*AC²=4----> AC²=1/2----> AC=√2/2
so
AB²=9*AC²----> AB²=9*(√2/2)²----> AB=(3√2)/2
the answer is
the hypotenuse is (3√2)/2
we know that
in the right triangle ABC
cos A=AC/AB
cos A=1/3
so
1/3=AC/AB----->AB=3*AC-----> square----> AB²=9*AC²----> equation 1
applying the Pythagoras Theorem
BC²+AC²=AB²-----> 2²+AC²=AB²---> 4+AC²=AB²----> equation 2
substitute equation 1 in equation 2
4+AC²=9*AC²----> 8*AC²=4----> AC²=1/2----> AC=√2/2
so
AB²=9*AC²----> AB²=9*(√2/2)²----> AB=(3√2)/2
the answer is
the hypotenuse is (3√2)/2
Answer:
a.
b.
c.
d.
Step-by-step explanation:
Given that there are two laptop machines and four desktop machines.
On a day, 2 computers to be set up.
To find:
a. probability that both selected setups are for laptop computers?
b. probability that both selected setups are desktop machines?
c. probability that at least one selected setup is for a desktop computer?
d. probability that at least one computer of each type is chosen for setup?
Solution:
Formula for probability of an event E can be observed as:
a. Favorable cases for Both the laptops to be selected = = 1
Total number of cases = 15
Required probability is .
b. Favorable cases for both the desktop machines selected =
Total number of cases = 15
Required probability is .
c. At least one desktop:
Two cases:
1. 1 desktop and 1 laptop:
Favorable cases =
2. Both desktop:
Favorable cases =
Total number of favorable cases = 8 + 6 = 14
Required probability is .
d. 1 desktop and 1 laptop:
Favorable cases =
Total number of cases = 15
Required probability is .