Answer: (I am really not sure the answer if it is wrong I am so sorry)
h=3, k=2
Horizontal: up, 2 units
Vertical: up, 3 units
Answer:
w=4
Step-by-step explanation:
-80/-5=4
Answer:
2 stickers will be left over
Step-by-step explanation:
To answer the question, all you need to do is figure out at what value of stickers would the remaining stickers be equal to 2 if the piles are made up of either 3 to a set or 4 to a set. To do this, we need to figure out the size of the whole set of stickers.
Lets assume that the total number of stickers is 14. Lets test the 2 given conditions. If we divide it in piles of 3 then we will have a pile of 12 stickers (4 piles of 3 stickers each) with 2 stickers remaining.
Lets test the second condition. If we divide it in piles of 4, we will again have 12 stickers (3 piles of 4 stickers each) with 2 stickers remaining.
Now that both conditions have been satisfied, all we have to do is see how many stickers are left if we make a pile of 12 stickers. In this case, we will have just one pile of 12 stickers and, again, we will be left with 2 stickers.
Therefore, the answer is 2 stickers
It is 23a+72
see
6(-3a+12)+8a-13a
-18a+72+8a-13a
23a+72
do you understand
Answer:
Test statistic = 1.818
Reject H0
Step-by-step explanation:
H0 : μ = 90
H1 : μ > 90
xbar = 94 ; s = 22 ; n = 100
The test statistic : (xbar - μ) ÷ (s/√(n))
Test statistic = (94 - 90) ÷ (22/√100)
Test statistic = 4 ÷ 2.2
T = 1.818
The critical value :
At α = 0.10
Degree of freedom = n - 1 = 100 - 1 = 99
Tcritical(0.10, 99) = 1.290
Decison region :
Reject H0 if Test statistic > |Tcritical |
1.818 > 1.290
We reject H0
