1. Exponents are the repeated multiplication of a number. It is represented
by the formula of a^n where a is the number to be repeated and n is the number
of times the number is being multiplies by itself. Also include the sign of the
number, no matter how it is multiplied, it is still important. For instance you have 4(4),
you have 4 as the repeated number. You can see that it is multiplied by itself
by two times. So you have (-4)^2.
2. Graphs and tables are the key features in finding a model between two quantities. If he data values, after plotting it into a graph, produces a straight line, then you will have a direct relationship between the two and sometimes you can get an equation of the line. Sometimes it will give you a curved line. That is why it is important to graph the values of the table to better understand the relationship between two variables.
Answer:
63 Jelly Beans
Step-by-step explanation:
The unknown is the number of jelly beans originally in the bag or x
First he had x jelly beans
The he ate one-third of them
He then ate 16 more jelly beans
This was equal to 37 jelly beans
This is the equation
Now solve for x
Subtract 16 from both sides
Multiply both sides by 3
63 Jelly Beans
Answer:
i think its b
Step-by-step explanation:
Answer:
the answer is A because if it is raised to the power greater than one and is added it will be greater than1
Answer:
(−0.103371 ; 0.063371) ;
No ;
( -0.0463642, 0.0063642)
Step-by-step explanation:
Shift 1:
Sample size, n1 = 30
Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm
Shift 2:
Sample size, n2 = 25
Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17
Mean difference ; μ1 - μ2
Zcritical at 95% confidence interval = 1.96
Using the relation :
(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)
(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)
Lower boundary :
-0.02 - 0.0833710 = −0.103371
Upper boundary :
-0.02 + 0.0833710 = 0.063371
(−0.103371 ; 0.063371)
B.)
We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.
C.)
For sample size :
n1 = 300 ; n2 = 250
(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)
Lower boundary :
-0.02 - 0.0263642 = −0.0463642
Upper boundary :
-0.02 + 0.0263642 = 0.0063642
( -0.0463642, 0.0063642)
