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!
Answer:
: Stroke Brain cells do not get the nutrients they need if blood flow to the brain …
Step-by-step explanation:
Answer:
[see below]
Step-by-step explanation:
I used a graphing program to graph the given system. I also graphed the points given.
Options B's choice lies in the double shaded area, so it would be a solution to the system.
See the graph attached!
Hope this helps you!
Answer:
I think 22.5
Step-by-step explanation:
okay so we know that lines bc and pf equal eachother
18/8 = 2.25 this is the scale factor
10 times 2.25 is. 22.5
I think it would be 22.5
Answer:
The equation is 
Step-by-step explanation:
The parameter that we have is t. We want to eliminate this parameter in both equations, therefore in the first equation we solve for t and in the second equation we solve for the variable t.
We have:
Now we solve the other equation for t.
<em> because</em> 
As 
and also 
Then:
