Answer:
A) an upper quartile of 56 and a median of 50
Step-by-step explanation:
Hello!
The lower and upper ends "box" from the box and whiskers plot is determined by the first (Q₁) and third (Q₃) quartiles of the data set and the line inside the box marks the median or second quartile of the sample (Q₂)
The length of the whiskers of the diagram is determined by the minimum value (left whisker) and the maximum value (right whisker)
If you observe the graphic:
Q₁= 42
Q₂= 50 (Median)
Q₃= 56
Min= 38
Max= 66
The box is delimited by a lower quartile of 42 and an upper quartile of 56, with a median 50, the least observation is 38 and the highest observation is 66.
The correct choice is A.
I hope this helps!
Answer: Domain: 0,1,2,3 and 4 and Range: 1,2,4,8 and 4
Step-by-step explanation:
The domain is the set of all inputs that makes the function definable. As we see the function looks like an exponential function which means the domain is all real numbers.
So the inputs are 0,1,2,3 and 4
the outputs are 1,2,4,8 and 4
Ok u have a phone and call 911 and say I don't know to solve can u help me
Answer:
The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes
Step-by-step explanation:
We have the standard deviation for the sample, but not for the population, so we use the students t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 35 - 1 = 35
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 34 degrees of freedom(y-axis) and a confidence level of ![1 - \frac{1 - 0.9}{2} = 0.975([tex]t_{975}](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7B1%20-%200.9%7D%7B2%7D%20%3D%200.975%28%5Btex%5Dt_%7B975%7D)
). So we have T = 2.0322
The margin of error is:
M = T*s = 2.0322*30 = 60.97
The upper end of the interval is the sample mean added to M. So it is 204 + 60.97 = 264.97
The upper boundary of the 95% confidence interval for the average unload time is 264.97 minutes
Time-and-a-half doesn't make enough sense to answer the question. If you fix it, or comment below the correction I will gladly help.
