yes sure, i learned this before so
1. a. make a proportion here, of percentage.
if you cross multiply and divide you get 90%.
1. b. obviously the empty seats would be 10%.
2. a. proportion again,
cross multiply then divide you ge 15%
3. 30% in decimal form is 0.3. multiply it then to the total amount, so 0.3 x 15 = 4.5
Jackson’s puppy would gain 80 pounds in 40 weeks
Answer:
54m^2
Step-by-step explanation:
Each square has a equal sides, which is 3m
in order to find the surface area, it's
Formula : Base x height = Area
The total surface area would be all the area's added together
3x3=9cm^2
so each square is 9m^2
there are 6 squares so
9+9+9+9+9+9=54m^2
or you can do 9x6=54m^2
Answer:
B= 90
C= 80
D= 100
Step-by-step explanation:
So let’s start with the bottom angles.
We know that the adjacent angles in a parallelogram add up to 180° so our equation would be
. So let’s simplify it b and b are both the same variables so we will get 2b and -10 + 10 is 0 so we get the following equation
.
So now we divide 180 and 2 and we get 90 as b.
So one side will be 100 because of the +10 and the other 90 because of the -10.
Now for the next side.
![d+c=180](https://tex.z-dn.net/?f=d%2Bc%3D180)
So we know that b+10 is adjacent to c so 180-100 is 80, so 80 is c.
Now for d.
b-10 is 80 so 180-80 is 100, so d is 100.
Answer: (20.86, 22.52)
Step-by-step explanation:
Formula to find the confidence interval for population mean :-
![\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%5Cpm%20z%5E%2A%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
, where
= sample mean.
z*= critical z-value
n= sample size.
= Population standard deviation.
By considering the given question , we have
![\overline{x}= 21.69](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D%2021.69)
![\sigma=3.23](https://tex.z-dn.net/?f=%5Csigma%3D3.23)
n= 58
Using z-table, the critical z-value for 95% confidence = z* = 1.96
Then, 95% confidence interval for the amount of time spent on administrative issues will be :
![21.69\pm (1.96)\dfrac{3.23}{\sqrt{58}}](https://tex.z-dn.net/?f=21.69%5Cpm%20%281.96%29%5Cdfrac%7B3.23%7D%7B%5Csqrt%7B58%7D%7D)
![=21.69\pm (1.96)\dfrac{1.7}{7.61577}](https://tex.z-dn.net/?f=%3D21.69%5Cpm%20%281.96%29%5Cdfrac%7B1.7%7D%7B7.61577%7D)
![=21.69\pm (1.96)(0.223221)](https://tex.z-dn.net/?f=%3D21.69%5Cpm%20%281.96%29%280.223221%29)
![\approx21.69\pm0.83](https://tex.z-dn.net/?f=%5Capprox21.69%5Cpm0.83)
![=(21.69-0.83,\ 21.69+0.83)=(20.86,\ 22.52)](https://tex.z-dn.net/?f=%3D%2821.69-0.83%2C%5C%2021.69%2B0.83%29%3D%2820.86%2C%5C%2022.52%29)
Hence, the 95% confidence interval for the amount of time spent on administrative issues = (20.86, 22.52)