(p - 0.15p) + 1.99
This is the equation to solve.
p minus 15% of p, plus 1.99 for shipping costs.
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:
14
Step-by-step explanation:
Arithmetic expressions are best evaluated using a reliable calculator. The Google calculator always follows the Order of Operations, so can give you a good answer to such questions.
(1/4)^-2 - (5^0)(2)(1^-1)
= 16 - (1)(2)(1) . . . . evaluate the exponents first
= 16 -2 . . . . . . . . . evaluate the product next
= 14 . . . . . . . . . . . finally, evaluate the sum
___
The rule of exponents is ...
a^-b = 1/(a^b)
Answer:
1) d ≤ -12
2) x ≥ -25
Step-by-step explanation:
-2.5d ≥ 30
2.5d ≤ -30
d ≤ -12
10 - x ≤ 35
-x ≤ 25
x ≥ -25
That is 12*11*10*9*8*7*6*5*4=79833600 ways
