Answer:
3.31 x 10^4
Step-by-step explanation:
3.111 x 10^4 is 31110
2 x 10^3 is 2000
31110 + 2000 = 33100 or 3.31 x 10^4
Your answer is B. y=-4x+2
Answer:
d) Convenience
Step-by-step explanation:
According to my research on different types of sampling techniques, I can say that based on the information provided within the question the sampling technique used in this method is called a Convenience technique. This technique collects data from all the participants that were conveniently available to participate, and does not have any criteria that needs to be met. In this situation they used the data from all the participants that responded to the questionnaires, in other words the participants that were conveniently available,
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
Actually Welcome to the concept of Parallel lines.
We must first understand that, Parallel Lines always have a same Slope, hence the 'm' value in y=mx+c equation will same, here it is '1/2' in the above equation,
so the points here are (-6,-17)
==>
(y-(-17)) = m(x-(-6))
==>
here m = 1/2 ,hence
y+17 = 1/2(x+6)
==> y+17 = 1/2(x) + 3
==> y = 1/2(x) + 3 - 17
==> y = 1/2(x) - 14
hence the Option 4.) is the correct answer!!
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!
