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
Answer:
The price of pretzels in 1975 = $1.80
Step-by-step explanation:
To answer this question we are assuming the steady rate means linear.Let x be the year and y be the price.
We need to find the slope
m = (y2-y1)/(x2-x1)
= (4.80-4.05)/(2015-2005)
=.75/10
= .075
The slope is .075
We can use the point slope form of the equation
y-y1 = m(x-x1)
y-4.80 = .075(x-2015)
Distribute
y - 4.80 = .075x -151.125
Add 4.80 to each side
y - 4.80+4.80 = .075x -151.125+4.80
y = .075 x - 146.325
We want to find out how much pretzels were in 1975. Put in x=1975
y = .075(1975) -146.325
y = 148.125-146.325
y=1.80
Step-by-step explanation:
(3X-20)+(3X+5)=111(( SINCE 111 IS TOTAL ANGLE
6X-15=111
6X=126
X=21
EFG=3×21-20
=63-20=43
<span>A)What is the maximum number of swimmers allowed on a swim team?</span>
Answer:
5.54 inches of rain
Step-by-step explanation:
Monday - 3.11
Tuesday - 0.81
Wednesday - 0.81 x 2= 1.62
3.11 + 0.81 + 1.62=5.54 inches
