♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫
➷The answer is that 20 goes into 16 1.25 times.
Explanation: You can use a calculator, and you just divide 20 by 16. Or, you can divide 20 and 16 both by 4. since you are dividing by four, and there are 5 fours in 20, you could know that 5*0.25 is 1.25, which is the answer. The fraction form is 1 1/4.
✽
➶ Hope This Helps You!
➶ Good Luck (:
➶ Have A Great Day ^-^
DOGE
If you use PEMDAS, you’ll start with multiplication. so 54*96=5,184. 6*2=12.
so now the problem would be 96+5,184+32/98-12.
from here you would divide.
so, 32/98=0.33 (rounded up)
now you add.
96+5,184+0.33=5,280.33
so, 5,280.33-12=5,268.33.
the answer is:
5,268.33
hope this helped!!
Answer:pretty sure its 10 degrees
Step-by-step explanation:
all tris add up to 180, and this one is a right angle so you know you have 90 deg, 80 degrees. 90+80=170; 180-170=10
Given Information:
Total cards = 108
Red cards = 25
yellow cards = 25
Blue cards = 25
Green cards = 25
Wild cards = 8
Required Information:
Probability that a hand will contain exactly two wild cards in a seven-hand game = ?
Answer:
P = (₈C₂*₁₀₀C₅)/₁₀₈C₇
Step-by-step explanation:
The required probability is given by
P = number of ways of interest/total number of ways
The total number of ways of dealing a seven-card hand is
₁₀₈C₇
We want to select exactly 2 wild cards and the total wild cards are 8 so the number of ways of this selection is
₈C₂
Since the game is seven-card hand, we have to get the number of ways to select remaining 5 cards out of (108 - 8 = 100) cards.
₁₀₀C₅
Therefore, the setup for this problem becomes
P = number of ways of interest/total number of ways
P = (₈C₂*₁₀₀C₅)/₁₀₈C₇
This is the required setup that we can type into our calculators to get the probability of exactly two wild cards in a seven-hand card game with 8 wild cards and 108 total cards.
Lets say that the vertical line in Y and Horizontal line in X
Take any point from either the number 8 diagonal or number 17 diagonal and see the value it has on both X and Y line and sum them , u will find the sum is always the same , this is because all the squares made by closest numbers r the same as any of the squares in this , think this as a graph paper that makes more sense if u r trying to understand ....
