Answer:
90% confidence interval for the true mean weight of orders is between a lower limit of 103.8645 grams and an upper limit of 116.1355 grams.
Step-by-step explanation:
Confidence interval for true mean weight is given as sample mean +/- margin of error (E)
sample mean = 110 g
sample sd = 14 g
n = 16
degree of freedom = n - 1 = 16 - 1 = 15
confidence level = 90% = 0.9
significance level = 1 - C = 1 - 0.9 = 0.1 = 10%
critical value (t) corresponding to 15 degrees of freedom and 10% significance level is 1.753
E = t × sample sd/√n = 1.753×14/√16 = 6.1355 g
Lower limit of sample mean = sample mean - E = 110 - 6.1355 = 103.8645 g
Upper limit of sample mean = sample mean + E = 110 + 6.1355 = 116.1355 g
90% confidence interval is (103.8645, 116.1355)
Firstly, let's create a function of f(t) where t represents the time that has past, and f(t) represents the amount of rainwater. We know that when t=1, then f(t)=10, and t=2 then f(t)=15. So, let's take that and analyze it:
(1,10)
(2,15)
m = (15-10)/(2-1) = 5
y-intercept = 5
∴ f(t) = 5t+5
Now we just evaluate t for 10:
f(10) = (5*10)+5
f(10) = 55
Answer:
1. 5/6 ≥ 2/3
2. 1/5 ≤ 2/8
3. 9/10 ≥ 6/8
4. I can't see it's out of the pic
5. 7/8 ≥ 5/10
6. 2/5 ≥ 2/6
7. 1/3 ≤ 3/8
8. I can't see it's out of the pic.
9. 8/10 ≥ 3/4
10. 3/8 ≤ 11/12
11. 2/3 ≤ 10/12
12. Can't see :(
13. 3/8 ≤ 7/8
14. 2/4 = 4/8
15. 6/8 ≥ 8/12
16. Can't see
Step-by-step explanation: All you have to do is divide the numerator by the denominator to make it into a decimal. For instance....
5/6 = approximately 0.83
I hope this helps you! It took forever....
here the equation of path is -2x-7=y
let us find slope of this line by using y=mx+b
so slope m=-2
now new path is perpendicular to this path.
product of slopes of perpendiculars is -1
so slope of new path is 1/2
now it passes through (-2,-3) with slope 1/2
using y=mx+b to get value of b
-3=1/2(-2) +b
b=-2
plugging m=-1/2 and b=-2 in y=mx + b to get the final equation as
y=(-1/2) x -2 is the equation of new path
Please comment on this answer and give me a full question and I will be just fine answering it for you.
