14b + 6b - 12
(14b + 6b) - 12 collect like terms
20b - 12 << your answer
hope this helps, God bless!
Answer:
(x,y)→(x+7,y)
Step-by-step explanation:
Cuz the point isn't moving up or down so the y is the same and x+7 cuz the point moved 7 units to the right so its position changed in the x-axis.
Im reallyyyy bad at explanations but hope this helped:)
Average speed=Total Distance Covered\<span>Total time take<span>n
T.D=240+180=420
T.T=2.5+2=4.5
420\4.5=93.3 km\h
answer=D</span></span>
Answer:
<u>25/7, 1.87, </u><u>3.25</u><u>,</u><u> </u><u>and</u><u> </u><u>2</u><u>2</u><u>/</u><u>7</u>
Step-by-step explanation:
16 < 15x - 12
__________
+12. +12
__________
28 < 15x
__________
÷15 ÷15
__________
28/15 < x
__________
<u>x > 28/15</u>
__________
this is true for any number greater than 1 and 13/15 = 1.8666......
Answer:
1) is not possible
2) P(A∪B) = 0.7
3) 1- P(A∪B) =0.3
4) a) C=A∩B' and P(C)= 0.3
b) P(D)= 0.4
Step-by-step explanation:
1) since the intersection of 2 events cannot be bigger than the smaller event then is not possible that P(A∩B)=0.5 since P(B)=0.4 . Thus the maximum possible value of P(A∩B) is 0.4
2) denoting A= getting Visa card , B= getting MasterCard the probability of getting one of the types of cards is given by
P(A∪B)= P(A)+P(B) - P(A∩B) = 0.6+0.4-0.3 = 0.7
P(A∪B) = 0.7
3) the probability that a student has neither type of card is 1- P(A∪B) = 1-0.7 = 0.3
4) the event C that the selected student has a visa card but not a MasterCard is given by C=A∩B' , where B' is the complement of B. Then
P(C)= P(A∩B') = P(A) - P(A∩B) = 0.6 - 0.3 = 0.3
the probability for the event D=a student has exactly one of the cards is
P(D)= P(A∩B') + P(A'∩B) = P(A∪B) - P(A∩B) = 0.7 - 0.3 = 0.4
