(x+6)^(1/2)-5=x+1
(x+6)^(1/2)=x+6
((x+6)^(1/2))^2=(x+6)^2
x+6=x^2+12x+36
0=x^2+11x+30
(-11+(11^2-4(1)(30))^(1/2))/2
(-11+((1)^(1/2))/2
(-11+1)/2=-5
(-11-1)/2=-6
((-6)+6)^1/2-5=(-6)+1
(0^(1/2))-5=-5
-6 is non-extraneous
((-5)+6)^1/2-5=(-5)+1
(1^1/2)-5=-4
1-5=-4
-4=-4
-5 is non-extraneous
Answer:
2.2gpa
Step-by-step explanation:
you add the value if the letter grades up and divide by how many classes you have. for this case you have 5 and you would add 4+3+2+1+0=10/5=2.2
Answer:
The value of the experimental probability is greater
Step-by-step explanation:
For the theoretical probability;
the probability that a card with the number 3 is selected is 1/5
We consider that the probabilities of each selection are equal, for the theoretical probability
For the experimental, we simply place the frequency of the selection 3 over the total
that will be 128/400 = 8/25
As we know that 8/25 is greater than 1/5
We can conclude that the value of the experimental probability is greater than that of the theoretical
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
