Answer:
We took $ 24 to the movies.
Step-by-step explanation:
Let total money be
On movies = Half of the money =![\frac{x}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B2%7D)
On popcorn = ![\frac{x}{4}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B4%7D)
on candy = $2
On soda = $3
total money spent =
+
+ 2+3+1
x =
+
+ 2+3+1
x-
-
= 6
-
-
=6
![\frac{4x-2x-x}{4} = 6\\ \frac{x}{4} =6\\x =24](https://tex.z-dn.net/?f=%5Cfrac%7B4x-2x-x%7D%7B4%7D%20%20%3D%206%5C%5C%20%20%20%20%20%20%20%20%20%20%5Cfrac%7Bx%7D%7B4%7D%20%3D6%5C%5Cx%20%3D24)
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
<span> x≥10
you add 4 to both sides</span>
Answer:
150
Step-by-step explanation
because 500 minus 350 is 150