2 times the square root of 5 . that’s what i got .
Answer:
a or c would be your answer hope this helps
Step-by-step explanation:
Answer:quotient property
Product property
Step-by-step explanation:
Just took the test
Answer:
Class Boundary = 1 between the sixth and seventh classes.
Step-by-step explanation:
Lengths (mm) Frequency
1. 140 - 143 1
2. 144 - 147 16
3. 148 - 151 71
4. 152 - 155 108
5. 156 - 159 83
6. 160 - 163 18
7. 164 - 167 3
The class boundary between two classes is the numerical value between the starting value of the higher class, which is 164 for the 7th class in this case, and the ending value of the class of the lower class, which is 163 for the 6th class in this case.
Therefore the class boundary between the sixth and seventh classes
= 164 - 163 = 1
Therefore Class Boundary = 1.
It can be seen that class boundary for the frequency distribution is 1.
If we take the difference between the lower limits of one class and the lower limit of the next class then we will get the class width value.
Therefore, Class width,
w = lower limit of second class - lower limit of first class
= 144 - 140
= 4
Answer:
10 grams or 0.010kg
Step-by-step explanation:
The mass of the bag used to make flour is 2.26 kg = 2260 grams
Now, in order to find out how many badges she made, we would need to divide the weight of the bag by the amount needed to make a badge.
Therefore: 
We know now that the bag will make 5 batches and will have some left over. In order to find how much we used, we can multiply 5 * 450 = 2250.
Therefore, in order to make 5 batches, Maria used exactly 2250 grams of flour which in result left us with 10 grams of the bag.
Note: 2260 -2250 = 10 grams.
Subtract, in order to find how many you have left.
Hope this helps.
