The answer is C) z - 2 = 5.
z - 2 = 5
Add 2 to both sides of the equal sign.
z - 2 + 2 = 5 + 2
z = 7.
Therefore, the solution is 7.
Hope this helps!
Answer:
10 + 12 = 24 so she can make 1..
Step-by-step explanation:
finding the volume of cube is equal to 2 × 3 × 4
so
v of cubes is 24
now
twice of it is 2 ×24
48
Answer:
Norton's
Step-by-step explanation:
Suppose that the down payment of Mazzeo's store = 1/3 on all installment purchases
Whereas Norton's depot required 30% down payment on installment purchases.
1/3 is fractional value of 33.33%.
therefore, Norton's Store's down payment rate is lower than that of Mazzeo's store.
At at least one die come up a 3?We can do this two ways:) The straightforward way is as follows. To get at least one 3, would be consistent with the following three mutually exclusive outcomes:the 1st die is a 3 and the 2nd is not: prob = (1/6)x(5/6)=5/36the 1st die is not a 3 and the 2nd is: prob = (5/6)x((1/6)=5/36both the 1st and 2nd come up 3: prob = (1/6)x(1/6)=1/36sum of the above three cases is prob for at least one 3, p = 11/36ii) A faster way is as follows: prob at least one 3 = 1 - (prob no 3's)The probability to get no 3's is (5/6)x(5/6) = 25/36.So the probability to get at least one 3 is, p = 1 - (25/36) = 11/362) What is the probability that a card drawn at random from an ordinary 52 deck of playing cards is a queen or a heart?There are 4 queens and 13 hearts, so the probability to draw a queen is4/52 and the probability to draw a heart is 13/52. But the probability to draw a queen or a heart is NOT the sum 4/52 + 13/52. This is because drawing a queen and drawing a heart are not mutually exclusive outcomes - the queen of hearts can meet both criteria! The number of cards which meet the criteria of being either a queen or a heart is only 16 - the 4 queens and the 12 remaining hearts which are not a queen. So the probability to draw a queen or a heart is 16/52 = 4/13.3) Five coins are tossed. What is the probability that the number of heads exceeds the number of tails?We can divide
