Answer:
Step-by-step explanation:
given that a deck of cards is shuffled.
we know in a deck there are 52 cards, 13 cards of each variety spade, clubs hearts and dice. Red are 26 and black are 26. kings, will be 4.
(a) the top card is the king of spades and the bottom card is the queen of spades?
(iii) 1/52 × 1/52
Top has 1/52 and bottom has 1/52 and these are independent.
(b) the top card is the king of spades and the bottom card is the king of spades?
(viii) None of the above
Because it is impossible.
(c) the top card is the king of spades or the bottom card is the king of spades?
(iv) 1/52 + 1/52
This is the sum of probabilities because there is no common event for these two.
(d) the top card is the king of spades or the bottom card is the queen of spades?
(ii) 1/52 + 1/51 (once king of spades is there, then probability is 1/51 for bottom card)
(e) of the top and bottom cards, one is the king of spades and the other is the queen of spades?
(vii) 2/52 × 1/51
Because this is twice of probability d.
Answer:
f = -5, s = -6
Step-by-step explanation:
F= first number
S= second number
4f - s = -14
f + 3s = -23
3(4f - s = -14)
12f - 3s = -42
+ f + 3s = -23
13f = -65
————
13
f = -5
4f - s = -14
4(-5) - s = -14
-20 - s = -14
+20 +20
-s = 6
———
-1
s = -6
Answer:
3/10
Step-by-step explanation:
7+3= 10 so 7+3=10/10 which is one mile
let me edit your question as:
Which two equations are true?
<u>Eq1:</u>
(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)
<u>Eq2:</u>
6.6×10−10(6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.7×103)
<u>Eq3:</u>
2.7×103(7.5×106)−(2.5×106)=5×100
Answer:
No one is true
Step-by-step explanation:
let's check each equation, if the values on both sides (left and right side) are equal then the equation is true otherwise false.
Using PEMDAS rule we are simplifying the equations as;
<u>Eq1:</u>
<u>Eq2:</u>
<u></u><u></u>
<u>Eq3:</u>
<u>we observed that none of the equation has two same values on both sides thus none of the three equations is true.</u>
<u>Also, no value of Eq1, Eq2 or Eq3 are same thus none of the equation is true</u>