The probability of one head and one tail is 2/3.
<u>Step-by-step explanation</u>:
- The possibilities for flipping two fair coins are {T,T}, {H,H}, {H,T}, {T,H}
- Given the case that at least one coin lands on a head, So the total possibilities are {H,H}, {H,T}, {T,H} = 3 possibilities
- Required event is 1 head and 1 tail= {H,T}, {T,H} = 2 possibilities
To calculate the probability of one head and one tail,
Probability = required events / Total events
Probability = 2/3
Answer:
18) Area= 5*5/2=25/2=12.5 unit ^2
19) Area=AB^2V3/4=8a^2*V3/4=2V3a^2
Step-by-step explanation:
18. A(-3,0)
B(1,-3)
C(4,1)
AB=V(-3-1)^2+(0+3)^2=V16+9=V25=5
AC=V(-3-4)^2+(0-1)^2=V49+1=V50=5V2
BC=V(1-4)^2+(-3-1)^2=V9+16=V25=5
so AB=BC=5
and AC^2=AB^2+BC^2
so trg ABC is an isosceles right angle triangle (<B=90)
Area= 5*5/2=25/2=12.5 unit ^2
19. A(a,a)
B(-a,-a)
C(-V3a, V3a)
AB=V(a+a)^2+(a+a)^2=V4a^2+4a^2=V8a^2
AC=V(a+V3a)^2+(a-V3)^2=Va^2+2a^2V3+3a^2+a^2-2a^2V3+3a^2=V8a^2
BC=V(-a+V3a)^2+(-a-V3a)^2=V8a^2
so AB=AC=BC
Area=AB^2V3/4=8a^2*V3/4=2V3a^2
Answer:
Matt ran 6 miles.
Step-by-step explanation:
Stan ran 4 and 7/10 miles, and this is 1 and 3/10 fewer miles than Matt.
this means that Matt ran the following amount of miles
4 and 7/|0 + 1 and 3/10 miles:
(4 + 1) + (7/10 + 3/10) = 5 + 10/10 = 6 miles.
This would be the correct way to solve this equation.
