Answer:
The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549
Step-by-step explanation:
We are in posessions of the sample's standard deviation. So we use the student's 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 = 20 - 1 = 19
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of ![1 - \frac{1 - 0.95}{2} = 0.975([tex]t_{975}](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7B1%20-%200.95%7D%7B2%7D%20%3D%200.975%28%5Btex%5Dt_%7B975%7D)
). So we have T = 2.0930
The margin of error is:
M = T*s = 2.0930*0.16 = 0.3349.
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 2.92 - 0.3349 = 2.5851
The upper end of the interval is the sample mean added to M. So it is 2.92 + 0.3349 = 3.2549
The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549
The answer is <span>$103.
To determine </span><span> how much Eileen will spend on gasoline, first we need to calculate how many gallons she needs.
If she drives 850 miles and </span><span>her car gets 23 miles per gallon, we can use the proportion:
850 miles : x gallons = 23 miles : 1 gallon
Crossing the products:
x = 850 miles </span>× 1 gallon ÷ 23 miles
x = 36.96 gallons
Thus, Eileen needs 36.96 gallons of gas. If the cost per gallon of gas is$2.79, using the proportion:
36.96 gallons : x = 1 gallon : <span>$2.79
x = 36.96 gallons </span>× $2.79 <span>÷ 1 gallon
x = </span>$103.12
x ≈ $<span>103</span>
2 inches wide
that means that the height of the large carton is 2 inches
find the volume of the carton
V=108 times volume of small box
V=108 times 1 times 2 time s3
V=648 in^2
height is 2
V=LWH
648=LW2 (note, this is not LW^2 but 2 times LW)
divie both sides by 2
324=LW
legnth of base is less than 36
div
we just need to find the values of legnth and width such that LW=324 and L<36 and that L or W is divisible by 3 (since each box is 3 inches long)
so lets factor 324
324=2*2*3*3*3*3
group them and find possible values for legnth and width less than 36 and divsible by 3
possible values
(L,W,H) possible values are
(27,12,2)
(18,18,2)
the possible dimentions are
27 by 12 by 2 or
18 by 18 by 2
Answer: The correct option is 1.
Explanation:
The given piecewise function is,
It means if x<0, then
If 
, then
Since the f(x) is defined for x<0 and , therefore the function f(x) is not defined for 
.
From the graph 2, 3 and 4 we can easily noticed that for each value of x there exist a unique value of y, therefore the function is defined for all values of x, which is not true according to the given piecewise function.
Only in figure the value of y not exist when x lies between 0 to 2, including 0. It means the function is not defined for , hence the first option is correct.
