Answer:
A. 4.4 I am doing good today
Answer:
10200
Step-by-step explanation:
This is an example of a compound interest. The equation for this problem is 10000*(1+2/100)^n. The n is the year as 1999 is 1 year away from 1998, you just replace n with 1 and you have a simple equation.
Ax^2 = bx Divide by x
ax^2/x = b
ax^(2 -1) = b
ax = b Divide by a
x = b/a
5 <<<< Answer.
x≠0
a≠0
Answer:
see below
Step-by-step explanation:
Any number between 650 and 749, 650 and 749 are included as well
if number is 50 and above it goes up
Answer:
Step-by-step explanation:
Researchers measured the data speeds for a particular smartphone carrier at 50 airports.
The highest speed measured was 76.6 Mbps.
n= 50
X[bar]= 17.95
S= 23.39
a. What is the difference between the carrier's highest data speed and the mean of all 50 data speeds?
If the highest speed is 76.6 and the sample mean is 17.95, the difference is 76.6-17.95= 58.65 Mbps
b. How many standard deviations is that [the difference found in part (a)]?
To know how many standard deviations is the max value apart from the sample mean, you have to divide the difference between those two values by the standard deviation
Dif/S= 58.65/23.39= 2.507 ≅ 2.51 Standard deviations
c. Convert the carrier's highest data speed to a z score.
The value is X= 76.6
Using the formula Z= (X - μ)/ δ= (76.6 - 17.95)/ 23.39= 2.51
d. If we consider data speeds that convert to z scores between minus−2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant?
The Z value corresponding to the highest data speed is 2.51, considerin that is greater than 2 you can assume that it is significant.
I hope it helps!
